Trắc nghiệm Phương trình lượng giác cơ bản Toán Lớp 11
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Câu 1:
Gọi \(x_0\) là nghiệm dương nhỏ nhất của \(\cos 2 x+\sqrt{3} \sin 2 x+\sqrt{3} \sin x-\cos x=2\) Mệnh đề nào sau đây là đúng?
A. \(x_{0} \in\left(0 ; \frac{\pi}{12}\right)\)
B. \(x_{0} \in\left[\frac{\pi}{12} ; \frac{\pi}{6}\right]\)
C. \(x_{0} \in\left(\frac{\pi}{6} ; \frac{\pi}{3}\right]\)
D. \(x_{0} \in\left(\frac{\pi}{3} ; \frac{\pi}{2}\right]\)
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Câu 2:
Gọi \(x_0\) là nghiệm âm lớn nhất của \(\sin 9 x+\sqrt{3} \cos 7 x=\sin 7 x+\sqrt{3} \cos 9 x\). Mệnh đề nào sau đây là đúng?
A. \(x_{0} \in\left(-\frac{\pi}{12} ; 0\right)\)
B. \(x_{0} \in\left[-\frac{\pi}{6} ;-\frac{\pi}{12}\right]\)
C. \(x_{0} \in\left[-\frac{\pi}{3} ;-\frac{\pi}{6}\right)\)
D. \(x_{0} \in\left[-\frac{\pi}{2} ;-\frac{\pi}{3}\right)\)
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Câu 3:
Giải phương trình \(\sqrt{3} \cos \left(x+\frac{\pi}{2}\right)+\sin \left(x-\frac{\pi}{2}\right)=2 \sin 2 x\)
A. \(\left[\begin{array}{l}x=\frac{5 \pi}{6}+k 2 \pi \\ x=\frac{\pi}{18}+k \frac{2 \pi}{3}\end{array}, k \in \mathbb{Z}\right.\)
B. \(\left[\begin{array}{l}x=\frac{7 \pi}{6}+k 2 \pi \\ x=-\frac{\pi}{18}+k \frac{2 \pi}{3}\end{array}, k \in \mathbb{Z}\right.\)
C. \(\left[\begin{array}{l}x=\frac{5 \pi}{6}+k 2 \pi \\ x=\frac{7 \pi}{6}+k 2 \pi\end{array}, k \in \mathbb{Z}\right.\)
D. \(\left[\begin{array}{l}x=\frac{\pi}{18}+k \frac{2 \pi}{3} \\ x=-\frac{\pi}{18}+k \frac{2 \pi}{3}\end{array}, k \in \mathbb{Z}\right.\)
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Câu 4:
Số nghiệm của phương trình \(\sin 5 x+\sqrt{3} \cos 5 x=2 \sin 7 x\) trên khoảng \(\left(0 ; \frac{\pi}{2}\right)\) là:
A. 1
B. 2
C. 3
D. 4
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Câu 5:
Tìm nghiệm dương nhỏ nhất của \(3 \sin 3 x-\sqrt{3} \cos 9 x=1+4 \sin ^{3} 3 x\)
A. \(x_{0}=\frac{\pi}{2}\)
B. \(x_{0}=\frac{\pi}{18}\)
C. \(x_{0}=\frac{\pi}{24}\)
D. \(x_{0}=\frac{\pi}{54}\)
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Câu 6:
Tính tổng T các nghiệm của phương trình \(\cos ^{2} x-\sin 2 x=\sqrt{2}+\sin ^{2} x\) trên khoảng \((0 ; 2 \pi)\)
A. \(T=\frac{7 \pi}{8}\)
B. \(T=\frac{21 \pi}{8}\)
C. \(T=\frac{11 \pi}{4}\)
D. \(T=\frac{3 \pi}{4}\)
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Câu 7:
Tổng hai nghiệm dương liên tiếp nhỏ nhất của phương trình \(\sin ^{6} x+\cos ^{6} x=\frac{7}{16}\) là:
A. \(\frac{5 \pi}{6}\)
B. \(\frac{\pi}{2}\)
C. \(\frac{7 \pi}{6}\)
D. \(\frac{\pi}{6}\)
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Câu 8:
Tổng nghiệm âm lớn nhất và nghiệm dương nhỏ nhất của phương trình \(\sin \left(3 x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{2}\) bằng:
A. \(\frac{\pi}{9}\)
B. \(-\frac{\pi}{6}\)
C. \(\frac{\pi}{6}\)
D. \(-\frac{\pi}{9}\)
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Câu 9:
Hỏi trên đoạn [-2017;2017], phương trình \((\sin x+1)(\sin x-\sqrt{2})=0\) có tất cả bao nhiêu nghiệm?
A. 4034.
B. 4035.
C. 641.
D. 642.
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Câu 10:
Tính đạo hàm của hàm số sau \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacI % cacaWG4bGaaiykaiabg2da9maaceaaeaqabeaacaWG4bWaaWbaaSqa % beaacaaIZaaaaOGaci4CaiaacMgacaGGUbWaaSaaaeaacaaIXaaaba % GaamiEaaaacaqGGaGaaeiiaiaabUgacaqGObGaaeyAaiaabccacaWG % 4bGaeyiyIKRaaGimaaqaaiaaicdacaqGGaGaaeiiaiaabccacaqGGa % GaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaae4AaiaabIga % caqGPbGaaeiiaiaadIhacqGH9aqpcaaIWaGaaeiiaaaacaGL7baaaa % a!57F9! f(x) = \left\{ \begin{array}{l} {x^3}\sin \frac{1}{x}{\rm{ khi }}x \ne 0\\ 0{\rm{ khi }}x = 0{\rm{ }} \end{array} \right.\)
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cacaGGOaGaamiEaiaacMcacqGH9aqpdaGabaabaeqabaGaamiEamaa % CaaaleqabaGaaGOmaaaakiGacohacaGGPbGaaiOBamaalaaabaGaaG % ymaaqaaiaadIhaaaGaeyOeI0IaamiEaiGacogacaGGVbGaai4Camaa % laaabaGaaGymaaqaaiaadIhaaaGaaeiiaiaabccacaqGGaGaae4Aai % aabIgacaqGPbGaaeiiaiaadIhacqGHGjsUcaaIWaaabaGaaeimaiaa % bccacaqGGaGaaeiiaiaabccacaqGGaGaae4AaiaabIgacaqGPbGaae % iiaiaabccacaqGGaGaamiEaiabg2da9iaaicdaaaGaay5Eaaaaaa!5D38! f'(x) = \left\{ \begin{array}{l} {x^2}\sin \frac{1}{x} - x\cos \frac{1}{x}{\rm{ khi }}x \ne 0\\ {\rm{0 khi }}x = 0 \end{array} \right.\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cacaGGOaGaamiEaiaacMcacqGH9aqpdaGabaabaeqabaGaaG4maiaa % dIhadaahaaWcbeqaaiaaikdaaaGcciGGZbGaaiyAaiaac6gadaWcaa % qaaiaaigdaaeaacaWG4baaaiabgkHiTiaadIhaciGGJbGaai4Baiaa % cohadaWcaaqaaiaaigdaaeaacaWG4baaaiaabccacaqGGaGaaeiiai % aabUgacaqGObGaaeyAaiaabccacaWG4bGaeyiyIKRaaGimaaqaaiaa % icdacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabUgacaqGObGaae % yAaiaabccacaqGGaGaaeiiaiaadIhacqGH9aqpcaaIWaaaaiaawUha % aaaa!5DFC! f'(x) = \left\{ \begin{array}{l} 3{x^2}\sin \frac{1}{x} - x\cos \frac{1}{x}{\rm{ khi }}x \ne 0\\ 0{\rm{ khi }}x = 0 \end{array} \right.\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cacaGGOaGaamiEaiaacMcacqGH9aqpdaGabaabaeqabaGaaG4maiaa % dIhadaahaaWcbeqaaiaaikdaaaGcciGGZbGaaiyAaiaac6gadaWcaa % qaaiaaigdaaeaacaWG4baaaiabgUcaRiaadIhaciGGJbGaai4Baiaa % cohadaWcaaqaaiaaigdaaeaacaWG4baaaiaabccacaqGGaGaaeiiai % aabUgacaqGObGaaeyAaiaabccacaWG4bGaeyiyIKRaaGimaaqaaiaa % icdacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabUgacaqGObGaae % yAaiaabccacaqGGaGaaeiiaiaadIhacqGH9aqpcaaIWaaaaiaawUha % aaaa!5DF1! f'(x) = \left\{ \begin{array}{l} 3{x^2}\sin \frac{1}{x} + x\cos \frac{1}{x}{\rm{ khi }}x \ne 0\\ 0{\rm{ khi }}x = 0 \end{array} \right.\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cacaGGOaGaamiEaiaacMcacqGH9aqpdaGabaabaeqabaGaaG4maiaa % dIhadaahaaWcbeqaaiaaikdaaaGcciGGZbGaaiyAaiaac6gadaWcaa % qaaiaaigdaaeaacaWG4baaaiabgkHiTiGacogacaGGVbGaai4Camaa % laaabaGaaGymaaqaaiaadIhaaaGaaeiiaiaabccacaqGGaGaae4Aai % aabIgacaqGPbGaaeiiaiaadIhacqGHGjsUcaaIWaaabaGaaGimaiaa % bccacaqGGaGaaeiiaiaabccacaqGGaGaae4AaiaabIgacaqGPbGaae % iiaiaabccacaqGGaGaamiEaiabg2da9iaaicdaaaGaay5Eaaaaaa!5CFF! f'(x) = \left\{ \begin{array}{l} 3{x^2}\sin \frac{1}{x} - \cos \frac{1}{x}{\rm{ khi }}x \ne 0\\ 0{\rm{ khi }}x = 0 \end{array} \right.\)
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Câu 11:
Tính đạo hàm của hàm số sau: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaaGymaaqaaiGacogacaGGVbGaai4CamaaCaaaleqa % baGaaGOmaaaakiaadIhacqGHsislciGGZbGaaiyAaiaac6gadaahaa % WcbeqaaiaaikdaaaGccaWG4baaaiabg2da9maalaaabaGaaGymaaqa % aiGacogacaGGVbGaai4CaiaaikdacaWG4baaaaaa!4998! y = \frac{1}{{{{\cos }^2}x - {{\sin }^2}x}} = \frac{1}{{\cos 2x}}\)
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaGkadaWcaa % qaaiGacohacaGGPbGaaiOBaiaaikdacaWG4baabaGaci4yaiaac+ga % caGGZbWaaWbaaSqabeaacaaIYaaaaOGaaGOmaiaadIhaaaGaaiOlaa % aa!4171! \frac{{\sin 2x}}{{{{\cos }^2}2x}}.\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaci % GGZbGaaiyAaiaac6gacaWG4baabaGaci4yaiaac+gacaGGZbWaaWba % aSqabeaacaaIYaaaaOGaaGOmaiaadIhaaaGaaiOlaaaa!400A! \frac{{\sin x}}{{{{\cos }^2}2x}}.\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIYaGaci4yaiaac+gacaGGZbGaaGOmaiaadIhaaeaaciGGZbGaaiyA % aiaac6gadaahaaWcbeqaaiaaikdaaaGccaaIYaGaamiEaaaacaGGUa % aaaa!4181! \frac{{2\cos 2x}}{{{{\sin }^2}2x}}.\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIYaGaci4CaiaacMgacaGGUbGaaGOmaiaadIhaaeaaciGGJbGaai4B % aiaacohadaahaaWcbeqaaiaaikdaaaGccaaIYaGaamiEaaaacaGGUa % aaaa!4182! \frac{{2\sin 2x}}{{{{\cos }^2}2x}}.\)
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Câu 12:
Tính đạo hàm của hàm số sau: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaci4CaiaacMgacaGGUbGaaGOmaiaadIhacqGHRaWk % ciGGJbGaai4BaiaacohacaaIYaGaamiEaaqaaiaaikdaciGGZbGaai % yAaiaac6gacaaIYaGaamiEaiabgkHiTiGacogacaGGVbGaai4Caiaa % ikdacaWG4baaaiaac6caaaa!4D7F! y = \frac{{\sin 2x + \cos 2x}}{{2\sin 2x - \cos 2x}}.\)
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI2aaabaWaaeWaaeaacaaIYaGaci4CaiaacMgacaGGUbGaaGOmaiaa % dIhacqGHsislciGGJbGaai4BaiaacohacaaIYaGaamiEaaGaayjkai % aawMcaamaaCaaaleqabaGaaGOmaaaaaaaaaa!43FC! \frac{6}{{{{\left( {2\sin 2x - \cos 2x} \right)}^2}}}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaaI2aaabaWaaeWaaeaaciGGZbGaaiyAaiaac6gacaaIYaGa % amiEaiabgkHiTiGacogacaGGVbGaai4CaiaaikdacaWG4baacaGLOa % GaayzkaaWaaWbaaSqabeaacaaIYaaaaaaaaaa!442D! \frac{{ - 6}}{{{{\left( {\sin 2x - \cos 2x} \right)}^2}}}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI2aaabaWaaeWaaeaacaaIYaGaci4CaiaacMgacaGGUbGaaGOmaiaa % dIhacqGHsislciGGJbGaai4BaiaacohacaWG4baacaGLOaGaayzkaa % WaaWbaaSqabeaacaaIYaaaaaaaaaa!4340! \frac{6}{{{{\left( {2\sin 2x - \cos x} \right)}^2}}}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaaI2aaabaWaaeWaaeaacaaIYaGaci4CaiaacMgacaGGUbGa % aGOmaiaadIhacqGHsislciGGJbGaai4BaiaacohacaaIYaGaamiEaa % GaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaaaaaa!44E9! \frac{{ - 6}}{{{{\left( {2\sin 2x - \cos 2x} \right)}^2}}}\)
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Câu 13:
Tính đạo hàm của hàm số sau: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacogacaGGVbGaai4CamaaCaaaleqabaGaaGOmaaaakmaabmaa % baWaaSaaaeaadaGcaaqaaiaadIhaaSqabaGccqGHRaWkcaaIXaaaba % WaaOaaaeaacaWG4baaleqaaOGaeyOeI0IaaGymaaaaaiaawIcacaGL % Paaaaaa!42E0! y = {\cos ^2}\left( {\frac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right)\)
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaigdaaeaadaGcaaqaaiaadIhaaSqabaGc % daqadaqaamaakaaabaGaamiEaaWcbeaakiabgkHiTiaaigdaaiaawI % cacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaOGaaiOlaiGacohacaGG % PbGaaiOBamaabmaabaWaaSaaaeaadaGcaaqaaiaadIhaaSqabaGccq % GHRaWkcaaIXaaabaWaaOaaaeaacaWG4baaleqaaOGaeyOeI0IaaGym % aaaaaiaawIcacaGLPaaacaGGUaaaaa!4B34! y' = \frac{1}{{\sqrt x {{\left( {\sqrt x - 1} \right)}^2}}}.\sin \left( {\frac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right).\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaigdaaeaadaGcaaqaaiaadIhaaSqabaGc % daqadaqaamaakaaabaGaamiEaaWcbeaakiabgkHiTiaaigdaaiaawI % cacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaOGaaiOlaiGacogacaGG % VbGaai4CamaabmaabaGaaGOmaiaac6cadaWcaaqaamaakaaabaGaam % iEaaWcbeaakiabgUcaRiaaigdaaeaadaGcaaqaaiaadIhaaSqabaGc % cqGHsislcaaIXaaaaaGaayjkaiaawMcaaiaac6caaaa!4C9C! y' = \frac{1}{{\sqrt x {{\left( {\sqrt x - 1} \right)}^2}}}.\cos \left( {2.\frac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right).\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaigdaaeaadaGcaaqaaiaadIhaaSqabaGc % daqadaqaamaakaaabaGaamiEaaWcbeaakiabgkHiTiaaigdaaiaawI % cacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaOGaaiOlaiGacohacaGG % PbGaaiOBamaabmaabaGaaGOmaiaac6cadaWcaaqaamaakaaabaGaam % iEaaWcbeaakiabgkHiTiaaigdaaeaadaGcaaqaaiaadIhaaSqabaGc % cqGHRaWkcaaIXaaaaaGaayjkaiaawMcaaiaac6caaaa!4CA2! y' = \frac{1}{{\sqrt x {{\left( {\sqrt x - 1} \right)}^2}}}.\sin \left( {2.\frac{{\sqrt x - 1}}{{\sqrt x + 1}}} \right).\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaigdaaeaadaGcaaqaaiaadIhaaSqabaGc % daqadaqaamaakaaabaGaamiEaaWcbeaakiabgkHiTiaaigdaaiaawI % cacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaOGaaiOlaiGacohacaGG % PbGaaiOBamaabmaabaGaaGOmaiaac6cadaWcaaqaamaakaaabaGaam % iEaaWcbeaakiabgUcaRiaaigdaaeaadaGcaaqaaiaadIhaaSqabaGc % cqGHsislcaaIXaaaaaGaayjkaiaawMcaaiaac6caaaa!4CA2! y' = \frac{1}{{\sqrt x {{\left( {\sqrt x - 1} \right)}^2}}}.\sin \left( {2.\frac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right).\)
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Câu 14:
Tính đạo hàm của hàm số sau: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maabmaabaWaaSaaaeaaciGGZbGaaiyAaiaac6gacaWG4baabaGa % aGymaiabgUcaRiGacogacaGGVbGaai4CaiaadIhaaaaacaGLOaGaay % zkaaWaaWbaaSqabeaacaaIZaaaaaaa!43BD! y = {\left( {\frac{{\sin x}}{{1 + \cos x}}} \right)^3}\)
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaci % GGZbGaaiyAaiaac6gadaahaaWcbeqaaiaaikdaaaGccaWG4baabaWa % aeWaaeaacaaIXaGaey4kaSIaci4yaiaac+gacaGGZbGaamiEaaGaay % jkaiaawMcaamaaCaaaleqabaGaaG4maaaaaaaaaa!42AC! \frac{{{{\sin }^2}x}}{{{{\left( {1 + \cos x} \right)}^3}}}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIZaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaIYaaaaOGaamiE % aaqaamaabmaabaGaaGymaiabgUcaRiGacogacaGGVbGaai4CaiaadI % haaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaaaa!4368! \frac{{3{{\sin }^2}x}}{{{{\left( {1 + \cos x} \right)}^2}}}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIYaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaIYaaaaOGaamiE % aaqaamaabmaabaGaaGymaiabgUcaRiGacogacaGGVbGaai4CaiaadI % haaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaaaa!4367! \frac{{2{{\sin }^2}x}}{{{{\left( {1 + \cos x} \right)}^2}}}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIZaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaIYaaaaOGaamiE % aaqaamaabmaabaGaaGymaiabgUcaRiGacogacaGGVbGaai4CaiaadI % haaiaawIcacaGLPaaadaahaaWcbeqaaiaaiodaaaaaaaaa!4369! \frac{{3{{\sin }^2}x}}{{{{\left( {1 + \cos x} \right)}^3}}}\)
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Câu 15:
Tính đạo hàm của hàm số sau \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maakeaabaGaamiEamaaCaaaleqabaGaaG4maaaakiabgUcaRiGa % cogacaGGVbGaai4CamaaCaaaleqabaGaaGinaaaakiaacIcacaaIYa % GaamiEaiabgkHiTmaalaaabaGaeqiWdahabaGaaG4maaaacaGGPaaa % leaacaaIZaaaaaaa!45F4! y = \sqrt[3]{{{x^3} + {{\cos }^4}(2x - \frac{\pi }{3})}}\)
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaiodacaWG4bWaaWbaaSqabeaacaaIYaaa % aOGaey4kaSIaaGioaiGacogacaGGVbGaai4CamaaCaaaleqabaGaaG % 4maaaakiaacIcacaaIYaGaamiEaiabgkHiTmaalaaabaGaeqiWdaha % baGaaGinaaaacaGGPaGaci4CaiaacMgacaGGUbGaaiikaiaaikdaca % WG4bGaeyOeI0YaaSaaaeaacqaHapaCaeaacaaI0aaaaiaacMcaaeaa % caaIZaWaaOqaaeaadaqadaqaaiaadIhadaahaaWcbeqaaiaaiodaaa % GccqGHRaWkciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaisdaaaGc % caGGOaGaaGOmaiaadIhacqGHsisldaWcaaqaaiabec8aWbqaaiaaio % daaaGaaiykaaGaayjkaiaawMcaamaaCaaaleqabaGaaG4maaaaaeaa % caaIZaaaaaaaaaa!61D8! y' = \frac{{3{x^2} + 8{{\cos }^3}(2x - \frac{\pi }{4})\sin (2x - \frac{\pi }{4})}}{{3\sqrt[3]{{{{\left( {{x^3} + {{\cos }^4}(2x - \frac{\pi }{3})} \right)}^3}}}}}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaiodacaWG4bWaaWbaaSqabeaacaaIYaaa % aOGaeyOeI0IaaGioaiGacogacaGGVbGaai4CamaaCaaaleqabaGaaG % 4maaaakiaacIcacaaIYaGaamiEaiabgkHiTmaalaaabaGaeqiWdaha % baGaaGinaaaacaGGPaGaci4CaiaacMgacaGGUbGaaiikaiaaikdaca % WG4bGaeyOeI0YaaSaaaeaacqaHapaCaeaacaaI0aaaaiaacMcaaeaa % caaI0aWaaOqaaeaadaqadaqaaiaadIhadaahaaWcbeqaaiaaiodaaa % GccqGHRaWkciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaisdaaaGc % caGGOaGaaGOmaiaadIhacqGHsisldaWcaaqaaiabec8aWbqaaiaaio % daaaGaaiykaaGaayjkaiaawMcaamaaCaaaleqabaGaaG4maaaaaeaa % caaIZaaaaaaaaaa!61E4! y' = \frac{{3{x^2} - 8{{\cos }^3}(2x - \frac{\pi }{4})\sin (2x - \frac{\pi }{4})}}{{4\sqrt[3]{{{{\left( {{x^3} + {{\cos }^4}(2x - \frac{\pi }{3})} \right)}^3}}}}}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaiAdacaWG4bWaaWbaaSqabeaacaaIYaaa % aOGaeyOeI0IaaGioaiGacogacaGGVbGaai4CamaaCaaaleqabaGaaG % 4maaaakiaacIcacaaIYaGaamiEaiabgkHiTmaalaaabaGaeqiWdaha % baGaaGinaaaacaGGPaGaci4CaiaacMgacaGGUbGaaiikaiaaikdaca % WG4bGaeyOeI0YaaSaaaeaacqaHapaCaeaacaaI0aaaaiaacMcaaeaa % caaIZaWaaOqaaeaadaqadaqaaiaadIhadaahaaWcbeqaaiaaiodaaa % GccqGHRaWkciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaisdaaaGc % caGGOaGaaGOmaiaadIhacqGHsisldaWcaaqaaiabec8aWbqaaiaaio % daaaGaaiykaaGaayjkaiaawMcaamaaCaaaleqabaGaaG4maaaaaeaa % caaIZaaaaaaaaaa!61E6! y' = \frac{{6{x^2} - 8{{\cos }^3}(2x - \frac{\pi }{4})\sin (2x - \frac{\pi }{4})}}{{3\sqrt[3]{{{{\left( {{x^3} + {{\cos }^4}(2x - \frac{\pi }{3})} \right)}^3}}}}}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaiodacaWG4bWaaWbaaSqabeaacaaIYaaa % aOGaeyOeI0IaaGioaiGacogacaGGVbGaai4CamaaCaaaleqabaGaaG % 4maaaakiaacIcacaaIYaGaamiEaiabgkHiTmaalaaabaGaeqiWdaha % baGaaGinaaaacaGGPaGaci4CaiaacMgacaGGUbGaaiikaiaaikdaca % WG4bGaeyOeI0YaaSaaaeaacqaHapaCaeaacaaI0aaaaiaacMcaaeaa % caaIZaWaaOqaaeaadaqadaqaaiaadIhadaahaaWcbeqaaiaaiodaaa % GccqGHRaWkciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaisdaaaGc % caGGOaGaaGOmaiaadIhacqGHsisldaWcaaqaaiabec8aWbqaaiaaio % daaaGaaiykaaGaayjkaiaawMcaamaaCaaaleqabaGaaG4maaaaaeaa % caaIZaaaaaaaaaa!61E3! y' = \frac{{3{x^2} - 8{{\cos }^3}(2x - \frac{\pi }{4})\sin (2x - \frac{\pi }{4})}}{{3\sqrt[3]{{{{\left( {{x^3} + {{\cos }^4}(2x - \frac{\pi }{3})} \right)}^3}}}}}\)
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Câu 16:
Tính đạo hàm của hàm số sau \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOmaaaakiaacIca % caaIZaGaamiEaiabgUcaRiaaigdacaGGPaaaaa!4073! y = {\sin ^2}(3x + 1)\)
A. 3sin(6x + 2)
B. sin(6x+2)
C. -3sin(6x + 2)
D. 3cos(6x+2)
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Câu 17:
Tính đạo hàm của hàm số sau \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maakaaabaGaaG4maiaadIhacqGHRaWkcaaIYaGaciiDaiaacgga % caGGUbGaamiEaaWcbeaaaaa!3F39! y = \sqrt {3x + 2\tan x} \)
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI1aGaey4kaSIaaGOmaiGacshacaGGHbGaaiOBamaaCaaaleqabaGa % aGOmaaaakiaadIhaaeaacaaIYaWaaOaaaeaacaaIZaGaamiEaiabgU % caRiaaikdaciGG0bGaaiyyaiaac6gacaWG4baaleqaaaaaaaa!451F! \frac{{5 + 2{{\tan }^2}x}}{{2\sqrt {3x + 2\tan x} }}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI1aGaeyOeI0IaaGOmaiGacshacaGGHbGaaiOBamaaCaaaleqabaGa % aGOmaaaakiaadIhaaeaacaaIYaWaaOaaaeaacaaIZaGaamiEaiabgU % caRiaaikdaciGG0bGaaiyyaiaac6gacaWG4baaleqaaaaaaaa!452A! \frac{{5 - 2{{\tan }^2}x}}{{2\sqrt {3x + 2\tan x} }}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaaI1aGaey4kaSIaaGOmaiGacshacaGGHbGaaiOBamaaCaaa % leqabaGaaGOmaaaakiaadIhaaeaacaaIYaWaaOaaaeaacaaIZaGaam % iEaiabgUcaRiaaikdaciGG0bGaaiyyaiaac6gacaWG4baaleqaaaaa % aaa!460C! \frac{{ - 5 + 2{{\tan }^2}x}}{{2\sqrt {3x + 2\tan x} }}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaaI1aGaeyOeI0IaaGOmaiGacshacaGGHbGaaiOBamaaCaaa % leqabaGaaGOmaaaakiaadIhaaeaacaaIYaWaaOaaaeaacaaIZaGaam % iEaiabgUcaRiaaikdaciGG0bGaaiyyaiaac6gacaWG4baaleqaaaaa % aaa!4617! \frac{{ - 5 - 2{{\tan }^2}x}}{{2\sqrt {3x + 2\tan x} }}\)
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Câu 18:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaci4yaiaac+gacaGGZbGaamiEaaqaaiaaikdaciGG % ZbGaaiyAaiaac6gadaahaaWcbeqaaiaaikdaaaGccaWG4baaaaaa!415C! y = \frac{{\cos x}}{{2{{\sin }^2}x}}\) có đạo hàm bằng:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0YaaS % aaaeaacaaIXaGaey4kaSIaci4CaiaacMgacaGGUbWaaWbaaSqabeaa % caaIYaaaaOGaamiEaaqaaiaaikdaciGGZbGaaiyAaiaac6gadaahaa % WcbeqaaiaaiodaaaGccaWG4baaaaaa!42DB! - \frac{{1 + {{\sin }^2}x}}{{2{{\sin }^3}x}}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0YaaS % aaaeaacaaIXaGaey4kaSIaci4yaiaac+gacaGGZbWaaWbaaSqabeaa % caaIYaaaaOGaamiEaaqaaiaaikdaciGGZbGaaiyAaiaac6gadaahaa % WcbeqaaiaaiodaaaGccaWG4baaaaaa!42D6! - \frac{{1 + {{\cos }^2}x}}{{2{{\sin }^3}x}}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaGaey4kaSIaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaIYaaa % aOGaamiEaaqaaiaaikdaciGGZbGaaiyAaiaac6gadaahaaWcbeqaai % aaiodaaaGccaWG4baaaaaa!41EE! \frac{{1 + {{\sin }^2}x}}{{2{{\sin }^3}x}}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaGaey4kaSIaci4yaiaac+gacaGGZbWaaWbaaSqabeaacaaIYaaa % aOGaamiEaaqaaiaaikdaciGGZbGaaiyAaiaac6gadaahaaWcbeqaai % aaiodaaaGccaWG4baaaaaa!41E9! \frac{{1 + {{\cos }^2}x}}{{2{{\sin }^3}x}}\)
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Câu 19:
Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadIhadaahaaWcbeqaaiaaikdaaaGcciGG0bGaaiyyaiaac6ga % caWG4bGaey4kaSYaaOaaaeaacaWG4baaleqaaaaa!3FAF! y = {x^2}\tan x + \sqrt x \) là
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIYaGaamiEaiGacshacaGGHbGaaiOBaiaadIhacqGH % RaWkdaWcaaqaaiaaigdaaeaacaaIYaWaaOaaaeaacaWG4baaleqaaa % aakiaac6caaaa!4266! y' = 2x\tan x + \frac{1}{{2\sqrt x }}.\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIYaaabaGaaG4maaaaaaa!377C! \frac{2}{3}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIYaGaamiEaiGacshacaGGHbGaaiOBaiaadIhacqGH % RaWkdaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaaakeaaciGGJb % Gaai4BaiaacohadaahaaWcbeqaaiaaikdaaaGccaWG4baaaiabgUca % RmaalaaabaGaaGymaaqaaiaaikdadaGcaaqaaiaadIhaaSqabaaaaO % GaaiOlaaaa!4A0B! y' = 2x\tan x + \frac{{{x^2}}}{{{{\cos }^2}x}} + \frac{1}{{2\sqrt x }}.\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIYaGaamiEaiGacshacaGGHbGaaiOBaiaadIhacqGH % RaWkdaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaaakeaaciGGJb % Gaai4BaiaacohadaahaaWcbeqaaiaaikdaaaGccaWG4baaaiabgUca % RmaalaaabaGaaGymaaqaamaakaaabaGaamiEaaWcbeaaaaGccaGGUa % aaaa!494F! y' = 2x\tan x + \frac{{{x^2}}}{{{{\cos }^2}x}} + \frac{1}{{\sqrt x }}.\)
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Câu 20:
Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacogacaGGVbGaaiiDamaaCaaaleqabaGaaGOmaaaakmaabmaa % baGaci4yaiaac+gacaGGZbGaamiEaaGaayjkaiaawMcaaiabgUcaRm % aakaaabaGaci4CaiaacMgacaGGUbGaamiEaiabgkHiTmaalaaabaGa % eqiWdahabaGaaGOmaaaaaSqabaaaaa!495F! y = {\cot ^2}\left( {\cos x} \right) + \sqrt {\sin x - \frac{\pi }{2}} \) là
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcqGHsislcaaIYaGaci4yaiaac+gacaGG0bWaaeWaaeaa % ciGGJbGaai4BaiaacohacaWG4baacaGLOaGaayzkaaWaaSaaaeaaca % aIXaaabaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaIYaaaaOWa % aeWaaeaaciGGJbGaai4BaiaacohacaWG4baacaGLOaGaayzkaaaaai % abgUcaRmaalaaabaGaci4yaiaac+gacaGGZbGaamiEaaqaaiaaikda % daGcaaqaaiGacohacaGGPbGaaiOBaiaadIhacqGHsisldaWcaaqaai % abec8aWbqaaiaaikdaaaaaleqaaaaakiaac6caaaa!5A07! y' = - 2\cot \left( {\cos x} \right)\frac{1}{{{{\sin }^2}\left( {\cos x} \right)}} + \frac{{\cos x}}{{2\sqrt {\sin x - \frac{\pi }{2}} }}.\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIYaGaci4yaiaac+gacaGG0bWaaeWaaeaaciGGJbGa % ai4BaiaacohacaWG4baacaGLOaGaayzkaaWaaSaaaeaacaaIXaaaba % Gaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaIYaaaaOWaaeWaaeaa % ciGGJbGaai4BaiaacohacaWG4baacaGLOaGaayzkaaaaaiaac6caci % GGZbGaaiyAaiaac6gacaWG4bGaey4kaSYaaSaaaeaaciGGJbGaai4B % aiaacohacaWG4baabaGaaGOmamaakaaabaGaci4CaiaacMgacaGGUb % GaamiEaiabgkHiTmaalaaabaGaeqiWdahabaGaaGOmaaaaaSqabaaa % aOGaaiOlaaaa!5DA1! y' = 2\cot \left( {\cos x} \right)\frac{1}{{{{\sin }^2}\left( {\cos x} \right)}}.\sin x + \frac{{\cos x}}{{2\sqrt {\sin x - \frac{\pi }{2}} }}.\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcqGHsislcaaIYaGaci4yaiaac+gacaGG0bWaaeWaaeaa % ciGGJbGaai4BaiaacohacaWG4baacaGLOaGaayzkaaWaaSaaaeaaca % aIXaaabaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaIYaaaaOWa % aeWaaeaaciGGJbGaai4BaiaacohacaWG4baacaGLOaGaayzkaaaaai % abgUcaRmaalaaabaGaci4yaiaac+gacaGGZbGaamiEaaqaamaakaaa % baGaci4CaiaacMgacaGGUbGaamiEaiabgkHiTmaalaaabaGaeqiWda % habaGaaGOmaaaaaSqabaaaaOGaaiOlaaaa!594B! y' = - 2\cot \left( {\cos x} \right)\frac{1}{{{{\sin }^2}\left( {\cos x} \right)}} + \frac{{\cos x}}{{\sqrt {\sin x - \frac{\pi }{2}} }}.\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIYaGaci4yaiaac+gacaGG0bWaaeWaaeaaciGGJbGa % ai4BaiaacohacaWG4baacaGLOaGaayzkaaWaaSaaaeaacaaIXaaaba % Gaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaIYaaaaOWaaeWaaeaa % ciGGJbGaai4BaiaacohacaWG4baacaGLOaGaayzkaaaaaiaac6caci % GGZbGaaiyAaiaac6gacaWG4bGaey4kaSYaaSaaaeaaciGGJbGaai4B % aiaacohacaWG4baabaWaaOaaaeaaciGGZbGaaiyAaiaac6gacaWG4b % GaeyOeI0YaaSaaaeaacqaHapaCaeaacaaIYaaaaaWcbeaaaaGccaGG % Uaaaaa!5CE5! y' = 2\cot \left( {\cos x} \right)\frac{1}{{{{\sin }^2}\left( {\cos x} \right)}}.\sin x + \frac{{\cos x}}{{\sqrt {\sin x - \frac{\pi }{2}} }}.\)
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Câu 21:
Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maakaaabaGaaGOmaiabgUcaRiGacshacaGGHbGaaiOBamaabmaa % baGaamiEaiabgUcaRmaalaaabaGaaGymaaqaaiaadIhaaaaacaGLOa % Gaayzkaaaaleqaaaaa!41B1! y = \sqrt {2 + \tan \left( {x + \frac{1}{x}} \right)} \) là
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmamaakaaabaGaaGOmaiab % gUcaRiGacshacaGGHbGaaiOBamaabmaabaGaamiEaiabgUcaRmaala % aabaGaaGymaaqaaiaadIhaaaaacaGLOaGaayzkaaaaleqaaaaakiab % gwSixdaa!4598! y' = \frac{1}{{2\sqrt {2 + \tan \left( {x + \frac{1}{x}} \right)} }} \cdot \)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0ZaaSaaaeaacaaIXaGaey4kaSIaciiDaiaacggacaGGUbWa % aWbaaSqabeaacaaIYaaaaOWaaeWaaeaacaWG4bGaey4kaSYaaSaaae % aacaaIXaaabaGaamiEaaaaaiaawIcacaGLPaaaaeaacaaIYaWaaOaa % aeaacaaIYaGaey4kaSIaciiDaiaacggacaGGUbWaaeWaaeaacaWG4b % Gaey4kaSYaaSaaaeaacaaIXaaabaGaamiEaaaaaiaawIcacaGLPaaa % aSqabaaaaOGaeyyXICnaaa!4F6E! y' = \frac{{1 + {{\tan }^2}\left( {x + \frac{1}{x}} \right)}}{{2\sqrt {2 + \tan \left( {x + \frac{1}{x}} \right)} }} \cdot \)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0ZaaSaaaeaacaaIXaGaey4kaSIaciiDaiaacggacaGGUbWa % aWbaaSqabeaacaaIYaaaaOWaaeWaaeaacaWG4bGaey4kaSYaaSaaae % aacaaIXaaabaGaamiEaaaaaiaawIcacaGLPaaaaeaacaaIYaWaaOaa % aeaacaaIYaGaey4kaSIaciiDaiaacggacaGGUbWaaeWaaeaacaWG4b % Gaey4kaSYaaSaaaeaacaaIXaaabaGaamiEaaaaaiaawIcacaGLPaaa % aSqabaaaaOGaaiOlamaabmaabaGaaGymaiabgkHiTmaalaaabaGaaG % ymaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaaaaaGccaGLOaGaayzk % aaGaaiOlaaaa!5474! y' = \frac{{1 + {{\tan }^2}\left( {x + \frac{1}{x}} \right)}}{{2\sqrt {2 + \tan \left( {x + \frac{1}{x}} \right)} }}.\left( {1 - \frac{1}{{{x^2}}}} \right).\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0ZaaSaaaeaacaaIXaGaey4kaSIaciiDaiaacggacaGGUbWa % aWbaaSqabeaacaaIYaaaaOWaaeWaaeaacaWG4bGaey4kaSYaaSaaae % aacaaIXaaabaGaamiEaaaaaiaawIcacaGLPaaaaeaacaaIYaWaaOaa % aeaacaaIYaGaey4kaSIaciiDaiaacggacaGGUbWaaeWaaeaacaWG4b % Gaey4kaSYaaSaaaeaacaaIXaaabaGaamiEaaaaaiaawIcacaGLPaaa % aSqabaaaaOGaaiOlamaabmaabaGaaGymaiabgUcaRmaalaaabaGaaG % ymaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaaaaaGccaGLOaGaayzk % aaGaaiOlaaaa!5469! y' = \frac{{1 + {{\tan }^2}\left( {x + \frac{1}{x}} \right)}}{{2\sqrt {2 + \tan \left( {x + \frac{1}{x}} \right)} }}.\left( {1 + \frac{1}{{{x^2}}}} \right).\)
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Câu 22:
Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOmaaaakmaabmaa % baWaaSaaaeaacqaHapaCaeaacaaIYaaaaiabgkHiTiaaikdacaWG4b % aacaGLOaGaayzkaaGaey4kaSYaaSaaaeaacqaHapaCaeaacaaIYaaa % aiaadIhacqGHsisldaWcaaqaaiabec8aWbqaaiaaisdaaaaaaa!4A5A! y = {\sin ^2}\left( {\frac{\pi }{2} - 2x} \right) + \frac{\pi }{2}x - \frac{\pi }{4}\) là
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaeyOeI0IaaGOmaiGacohacaGGPbGaaiOBamaabmaabaGa % eqiWdaNaeyOeI0IaaGinaiaadIhaaiaawIcacaGLPaaacqGHRaWkda % Wcaaqaaiabec8aWbqaaiaaikdaaaGaeyyXICnaaa!4827! y' = - 2\sin \left( {\pi - 4x} \right) + \frac{\pi }{2} \cdot \)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGOmaiGacohacaGGPbGaaiOBamaabmaabaWaaSaaaeaa % cqaHapaCaeaacaaIYaaaaiabgkHiTiaadIhaaiaawIcacaGLPaaaci % GGJbGaai4BaiaacohadaqadaqaamaalaaabaGaeqiWdahabaGaaGOm % aaaacqGHsislcaWG4baacaGLOaGaayzkaaGaey4kaSYaaSaaaeaacq % aHapaCaeaacaaIYaaaaiaac6caaaa!4E7F! y' = 2\sin \left( {\frac{\pi }{2} - x} \right)\cos \left( {\frac{\pi }{2} - x} \right) + \frac{\pi }{2}.\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGOmaiGacohacaGGPbGaaiOBamaabmaabaWaaSaaaeaa % cqaHapaCaeaacaaIYaaaaiabgkHiTiaadIhaaiaawIcacaGLPaaaci % GGJbGaai4BaiaacohadaqadaqaamaalaaabaGaeqiWdahabaGaaGOm % aaaacqGHsislcaWG4baacaGLOaGaayzkaaGaey4kaSYaaSaaaeaacq % aHapaCaeaacaaIYaaaaiaadIhacaGGUaaaaa!4F7C! y' = 2\sin \left( {\frac{\pi }{2} - x} \right)\cos \left( {\frac{\pi }{2} - x} \right) + \frac{\pi }{2}x.\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaeyOeI0IaaGOmaiGacohacaGGPbGaaiOBamaabmaabaGa % eqiWdaNaeyOeI0IaaGinaiaadIhaaiaawIcacaGLPaaacaGGUaaaaa!4324! y' = - 2\sin \left( {\pi - 4x} \right).\)
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Câu 23:
Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iabgkHiTmaalaaabaGaaGOmaaqaaiGacshacaGGHbGaaiOBamaa % bmaabaGaaGymaiabgkHiTiaaikdacaWG4baacaGLOaGaayzkaaaaaa % aa!416C! y = - \frac{2}{{\tan \left( {1 - 2x} \right)}}\) bằng:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI0aGaamiEaaqaaiGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOm % aaaakmaabmaabaGaaGymaiabgkHiTiaaikdacaWG4baacaGLOaGaay % zkaaaaaaaa!4074! \frac{{4x}}{{{{\sin }^2}\left( {1 - 2x} \right)}}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaaI0aaabaGaci4CaiaacMgacaGGUbWaaeWaaeaacaaIXaGa % eyOeI0IaaGOmaiaadIhaaiaawIcacaGLPaaaaaaaaa!3F71! \frac{{ - 4}}{{\sin \left( {1 - 2x} \right)}}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaaI0aGaamiEaaqaaiGacohacaGGPbGaaiOBamaaCaaaleqa % baGaaGOmaaaakmaabmaabaGaaGymaiabgkHiTiaaikdacaWG4baaca % GLOaGaayzkaaaaaaaa!4161! \frac{{ - 4x}}{{{{\sin }^2}\left( {1 - 2x} \right)}}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaaI0aaabaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaI % YaaaaOWaaeWaaeaacaaIXaGaeyOeI0IaaGOmaiaadIhaaiaawIcaca % GLPaaaaaaaaa!4064! \frac{{ - 4}}{{{{\sin }^2}\left( {1 - 2x} \right)}}\)
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Câu 24:
Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgacaGGOaGaamiEaiaacMcacqGHsislciGGJbGaai4Baiaa % cohadaahaaWcbeqaaiaaikdaaaGccaWG4baaaa!40E9! y = f(x) - {\cos ^2}x\) với f(x) là hàm liên tục trên R. Trong bốn biểu thức dưới đây, biểu thức nào xác định hàm f(x) thỏa mãn y’=1 với mọi \(x\in R\)?
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWG4bGaey4kaSYaaSaaaeaacaaIXaaabaGaaGOmaaaaciGGJbGa % ai4BaiaacohacaaIYaGaamiEaaaa!3E05! x + \frac{1}{2}\cos 2x\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgk % HiTmaalaaabaGaaGymaaqaaiaaikdaaaGaci4yaiaac+gacaGGZbGa % aGOmaiaadIhaaaa!3DF1! x - \frac{1}{2}\cos 2x\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWG4bGaeyOeI0Iaci4CaiaacMgacaGGUbGaaGOmaiaadIhaaaa!3C8E! x - \sin 2x\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWG4bGaeyOeI0Iaci4CaiaacMgacaGGUbGaaGOmaiaadIhaaaa!3C8E! x + \sin 2x\)
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Câu 25:
Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacshacaGGHbGaaiOBamaaCaaaleqabaGaaGOmaaaakiaadIha % cqGHsislciGGJbGaai4BaiaacshadaahaaWcbeqaaiaaikdaaaGcca % WG4baaaa!4269! y = {\tan ^2}x - {\cot ^2}x\) là
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGOmamaalaaabaGaciiDaiaacggacaGGUbGaamiEaaqa % aiGacogacaGGVbGaai4CamaaCaaaleqabaGaaGOmaaaakiaadIhaaa % Gaey4kaSIaaGOmamaalaaabaGaci4yaiaac+gacaGG0bGaamiEaaqa % aiGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOmaaaakiaadIhaaa % GaeyyXICnaaa!4DF1! y' = 2\frac{{\tan x}}{{{{\cos }^2}x}} + 2\frac{{\cot x}}{{{{\sin }^2}x}} \cdot \)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGOmamaalaaabaGaciiDaiaacggacaGGUbGaamiEaaqa % aiGacogacaGGVbGaai4CamaaCaaaleqabaGaaGOmaaaakiaadIhaaa % GaeyOeI0IaaGOmamaalaaabaGaci4yaiaac+gacaGG0bGaamiEaaqa % aiGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOmaaaakiaadIhaaa % GaeyyXICnaaa!4DFC! y' = 2\frac{{\tan x}}{{{{\cos }^2}x}} - 2\frac{{\cot x}}{{{{\sin }^2}x}} \cdot \)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGOmamaalaaabaGaciiDaiaacggacaGGUbGaamiEaaqa % aiGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOmaaaakiaadIhaaa % Gaey4kaSIaaGOmamaalaaabaGaci4yaiaac+gacaGG0bGaamiEaaqa % aiGacogacaGGVbGaai4CamaaCaaaleqabaGaaGOmaaaakiaadIhaaa % GaeyyXICnaaa!4DF1! y' = 2\frac{{\tan x}}{{{{\sin }^2}x}} + 2\frac{{\cot x}}{{{{\cos }^2}x}} \cdot \)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGOmaiGacshacaGGHbGaaiOBaiaadIhacqGHsislcaaI % YaGaci4yaiaac+gacaGG0bGaamiEaiaac6caaaa!42B9! y' = 2\tan x - 2\cot x.\)
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Câu 26:
Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOmaaaakiaaikda % caWG4bGaaiOlaiGacogacaGGVbGaai4CaiaadIhacqGHRaWkdaWcaa % qaaiaaikdaaeaadaGcaaqaaiaadIhaaSqabaaaaaaa!44C3! y = {\sin ^2}2x.\cos x + \frac{2}{{\sqrt x }}\) là
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGOmaiGacohacaGGPbGaaiOBaiaaikdacaWG4bGaaiOl % aiGacogacaGGVbGaai4CaiaadIhacqGHsislciGGZbGaaiyAaiaac6 % gacaWG4bGaaiOlaiGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOm % aaaakiaaikdacaWG4bGaeyOeI0IaaGOmamaakaaabaGaamiEaaWcbe % aakiaac6caaaa!5047! y' = 2\sin 2x.\cos x - \sin x.{\sin ^2}2x - 2\sqrt x .\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGOmaiGacohacaGGPbGaaiOBaiaaikdacaWG4bGaaiOl % aiGacogacaGGVbGaai4CaiaadIhacqGHsislciGGZbGaaiyAaiaac6 % gacaWG4bGaaiOlaiGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOm % aaaakiaaikdacaWG4bGaeyOeI0IaaGOmamaakaaabaGaamiEaaWcbe % aakiaac6caaaa!5047! y' = 2\sin 2x.\cos x - \sin x.{\sin ^2}2x - 2\sqrt x .\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGOmaiGacohacaGGPbGaaiOBaiaaisdacaWG4bGaaiOl % aiGacogacaGGVbGaai4CaiaadIhacqGHRaWkciGGZbGaaiyAaiaac6 % gacaWG4bGaaiOlaiGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOm % aaaakiaaikdacaWG4bGaeyOeI0YaaSaaaeaacaaIXaaabaGaamiEam % aakaaabaGaamiEaaWcbeaaaaGccqGHflY1aaa!52E2! y' = 2\sin 4x.\cos x + \sin x.{\sin ^2}2x - \frac{1}{{x\sqrt x }} \cdot \)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGOmaiGacohacaGGPbGaaiOBaiaaisdacaWG4bGaaiOl % aiGacogacaGGVbGaai4CaiaadIhacqGHsislciGGZbGaaiyAaiaac6 % gacaWG4bGaaiOlaiGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOm % aaaakiaaikdacaWG4bGaeyOeI0YaaSaaaeaacaaIXaaabaGaamiEam % aakaaabaGaamiEaaWcbeaaaaGccqGHflY1aaa!52ED! y' = \sin 4x.\cos x - \sin x.{\sin ^2}2x - \frac{1}{{x\sqrt x }} \cdot \)
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Câu 27:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maakaaabaGaci4yaiaac+gacaGG0bGaaGOmaiaadIhaaSqabaaa % aa!3CA0! y = \sqrt {\cot 2x} \) có đạo hàm là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaigdacqGHRaWkciGGJbGaai4Baiaacsha % daahaaWcbeqaaiaaikdaaaGccaaIYaGaamiEaaqaamaakaaabaGaci % 4yaiaac+gacaGG0bGaaGOmaiaadIhaaSqabaaaaaaa!4478! y' = \frac{{1 + {{\cot }^2}2x}}{{\sqrt {\cot 2x} }}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiabgkHiTmaabmaabaGaaGymaiabgUcaRiGa % cogacaGGVbGaaiiDamaaCaaaleqabaGaaGOmaaaakiaaikdacaWG4b % aacaGLOaGaayzkaaaabaWaaOaaaeaaciGGJbGaai4BaiaacshacaaI % YaGaamiEaaWcbeaaaaaaaa!46EE! y' = \frac{{ - \left( {1 + {{\cot }^2}2x} \right)}}{{\sqrt {\cot 2x} }}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaigdacqGHRaWkciGG0bGaaiyyaiaac6ga % daahaaWcbeqaaiaaikdaaaGccaaIYaGaamiEaaqaamaakaaabaGaci % 4yaiaac+gacaGG0bGaaGOmaiaadIhaaSqabaaaaaaa!4475! y' = \frac{{1 + {{\tan }^2}2x}}{{\sqrt {\cot 2x} }}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiabgkHiTmaabmaabaGaaGymaiabgUcaRiGa % cshacaGGHbGaaiOBamaaCaaaleqabaGaaGOmaaaakiaaikdacaWG4b % aacaGLOaGaayzkaaaabaWaaOaaaeaaciGGJbGaai4BaiaacshacaaI % YaGaamiEaaWcbeaaaaaaaa!46EA! y' = \frac{{ - \left( {1 + {{\tan }^2}2x} \right)}}{{\sqrt {\cot 2x} }}\)
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Câu 28:
Tính đạo hàm của hàm số sau: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maabmaabaGaci4yaiaac+gacaGGZbWaaWbaaSqabeaacaaI0aaa % aOGaamiEaiabgkHiTiGacohacaGGPbGaaiOBamaaCaaaleqabaGaaG % inaaaakiaadIhaaiaawIcacaGLPaaadaahaaWcbeqaaiaaiwdaaaaa % aa!44E9! y = {\left( {{{\cos }^4}x - {{\sin }^4}x} \right)^5}\)
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % ymaiaaicdaciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaisdaaaGc % caaIYaGaamiEaiaac6caaaa!3E89! - 10{\cos ^4}2x.\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Iaci % 4yaiaac+gacaGGZbWaaWbaaSqabeaacaaI0aaaaOGaaGOmaiaadIha % caGGUaGaci4CaiaacMgacaGGUbGaaGOmaiaadIhacaGGUaaaaa!4257! - {\cos ^4}2x.\sin 2x.\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % ymaiaaicdaciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaisdaaaGc % caaIYaGaamiEaiaac6caciGGZbGaaiyAaiaac6gacaWG4bGaaiOlaa % aa!4310! - 10{\cos ^4}2x.\sin x.\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % ymaiaaicdaciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaisdaaaGc % caaIYaGaamiEaiaac6caciGGZbGaaiyAaiaac6gacaaIYaGaamiEai % aac6caaaa!43CC! - 10{\cos ^4}2x.\sin 2x.\)
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Câu 29:
Tính đạo hàm của hàm số sau: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maabmaabaGaci4CaiaacMgacaGGUbGaamiEaiabgUcaRiGacoga % caGGVbGaai4CaiaadIhaaiaawIcacaGLPaaadaahaaWcbeqaaiaaio % daaaaaaa!42F2! y = {\left( {\sin x + \cos x} \right)^3}\)
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaG4mamaabm % aabaGaci4CaiaacMgacaGGUbGaamiEaiabgUcaRiGacogacaGGVbGa % ai4CaiaadIhaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGcda % qadaqaaiGacogacaGGVbGaai4CaiaadIhacqGHRaWkciGGZbGaaiyA % aiaac6gacaWG4baacaGLOaGaayzkaaGaaiOlaaaa!4C76! 3{\left( {\sin x + \cos x} \right)^2}\left( {\cos x + \sin x} \right).\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaG4mamaabm % aabaGaci4CaiaacMgacaGGUbGaamiEaiabgkHiTiaadogacaGGVbGa % ai4CaiaadIhaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGcda % qadaqaaiGacogacaGGVbGaai4CaiaadIhacqGHsislciGGZbGaaiyA % aiaac6gacaWG4baacaGLOaGaayzkaaGaaiOlaaaa!4C8B! 3{\left( {\sin x - cosx} \right)^2}\left( {\cos x - \sin x} \right).\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaci % GGZbGaaiyAaiaac6gacaWG4bGaey4kaSIaci4yaiaac+gacaGGZbGa % amiEaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakmaabmaaba % Gaci4yaiaac+gacaGGZbGaamiEaiabgkHiTiGacohacaGGPbGaaiOB % aiaadIhaaiaawIcacaGLPaaacaGGUaaaaa!4BC4! {\left( {\sin x + \cos x} \right)^2}\left( {\cos x - \sin x} \right).\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaG4mamaabm % aabaGaci4CaiaacMgacaGGUbGaamiEaiabgUcaRiGacogacaGGVbGa % ai4CaiaadIhaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGcda % qadaqaaiGacogacaGGVbGaai4CaiaadIhacqGHsislciGGZbGaaiyA % aiaac6gacaWG4baacaGLOaGaayzkaaGaaiOlaaaa!4C81! 3{\left( {\sin x + \cos x} \right)^2}\left( {\cos x - \sin x} \right).\)
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Câu 30:
Tính đạo hàm của hàm số sau: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maakaaabaGaci4CaiaacMgacaGGUbGaamiEaiabgUcaRiaaikda % caWG4baaleqaaaaa!3E83! y = \sqrt {\sin x + 2x} \)
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaci % GGJbGaai4BaiaacohacaWG4bGaey4kaSIaaGOmaaqaaiaaikdadaGc % aaqaaiGacohacaGGPbGaaiOBaiaadIhacqGHRaWkcaaIYaGaamiEaa % WcbeaaaaGccaGGUaaaaa!4375! \frac{{\cos x + 2}}{{2\sqrt {\sin x + 2x} }}.\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaci % GGJbGaai4BaiaacohacaWG4bGaey4kaSIaaGOmaaqaamaakaaabaGa % ci4CaiaacMgacaGGUbGaamiEaiabgUcaRiaaikdacaWG4baaleqaaa % aakiaac6caaaa!42B9! \frac{{\cos x + 2}}{{\sqrt {\sin x + 2x} }}.\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIYaaabaGaaGOmamaakaaabaGaci4CaiaacMgacaGGUbGaamiEaiab % gUcaRiaaikdacaWG4baaleqaaaaakiaac6caaaa!3EC3! \frac{2}{{2\sqrt {\sin x + 2x} }}.\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaci % GGJbGaai4BaiaacohacaWG4baabaGaaGOmamaakaaabaGaci4Caiaa % cMgacaGGUbGaamiEaiabgUcaRiaaikdacaWG4baaleqaaaaakiaac6 % caaaa!41D7! \frac{{\cos x}}{{2\sqrt {\sin x + 2x} }}.\)
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Câu 31:
Tính đạo hàm của hàm số sau \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacohacaGGPbGaaiOBamaakaaabaGaaGOmaiabgUcaRiaadIha % daahaaWcbeqaaiaaikdaaaaabeaaaaa!3E64! y = \sin \sqrt {2 + {x^2}} \)
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4yaiaac+ % gacaGGZbWaaOaaaeaacaaIYaGaey4kaSIaamiEamaaCaaaleqabaGa % aGOmaaaaaeqaaOGaaiOlaaaa!3D17! \cos \sqrt {2 + {x^2}} .\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaWaaOaaaeaacaaIYaGaey4kaSIaamiEamaaCaaaleqabaGa % aGOmaaaaaeqaaaaakiaac6caciGGJbGaai4BaiaacohadaGcaaqaai % aaikdacqGHRaWkcaWG4bWaaWbaaSqabeaacaaIYaaaaaqabaGccaGG % Uaaaaa!4232! \frac{1}{{\sqrt {2 + {x^2}} }}.\cos \sqrt {2 + {x^2}} .\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaaGOmaaaacaGGUaGaci4yaiaac+gacaGGZbWaaOaaaeaa % caaIYaGaey4kaSIaamiEamaaCaaaleqabaGaaGOmaaaaaeqaaOGaai % Olaaaa!3F50! \frac{1}{2}.\cos \sqrt {2 + {x^2}} .\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4baabaWaaOaaaeaacaaIYaGaey4kaSIaamiEamaaCaaaleqabaGa % aGOmaaaaaeqaaaaakiaac6caciGGJbGaai4BaiaacohadaGcaaqaai % aaikdacqGHRaWkcaWG4bWaaWbaaSqabeaacaaIYaaaaaqabaGccaGG % Uaaaaa!4274! \frac{x}{{\sqrt {2 + {x^2}} }}.\cos \sqrt {2 + {x^2}} .\)
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Câu 32:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacshacaGGHbGaaiOBamaaCaaaleqabaGaaGOmaaaakmaalaaa % baGaamiEaaqaaiaaikdaaaaaaa!3D85! y = {\tan ^2}\frac{x}{2}\) có đạo hàm là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiGacohacaGGPbGaaiOBamaalaaabaGaamiE % aaqaaiaaikdaaaaabaGaci4yaiaac+gacaGGZbWaaWbaaSqabeaaca % aIZaaaaOWaaSaaaeaacaWG4baabaGaaGOmaaaaaaaaaa!42E4! y' = \frac{{\sin \frac{x}{2}}}{{{{\cos }^3}\frac{x}{2}}}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaikdaciGGZbGaaiyAaiaac6gadaWcaaqa % aiaadIhaaeaacaaIYaaaaaqaaiGacogacaGGVbGaai4CamaaCaaale % qabaGaaG4maaaakmaalaaabaGaamiEaaqaaiaaikdaaaaaaaaa!43A0! y' = \frac{{2\sin \frac{x}{2}}}{{{{\cos }^3}\frac{x}{2}}}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiGacohacaGGPbGaaiOBamaalaaabaGaamiE % aaqaaiaaikdaaaaabaGaaGOmaiGacogacaGGVbGaai4CamaaCaaale % qabaGaaG4maaaakmaalaaabaGaamiEaaqaaiaaikdaaaaaaaaa!43A0! y' = \frac{{\sin \frac{x}{2}}}{{2{{\cos }^3}\frac{x}{2}}}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpciGG0bGaaiyyaiaac6gadaahaaWcbeqaaiaaiodaaaGc % daqadaqaamaalaaabaGaamiEaaqaaiaaikdaaaaacaGLOaGaayzkaa % aaaa!3FB9! y' = {\tan ^3}\left( {\frac{x}{2}} \right)\)
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Câu 33:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaaikdadaGcaaqaaiGacohacaGGPbGaaiOBaiaadIhaaSqabaGc % cqGHsislcaaIYaWaaOaaaeaaciGGJbGaai4BaiaacohacaWG4baale % qaaaaa!4242! y = 2\sqrt {\sin x} - 2\sqrt {\cos x} \) có đạo hàm là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaigdaaeaadaGcaaqaaiGacohacaGGPbGa % aiOBaiaadIhaaSqabaaaaOGaeyOeI0YaaSaaaeaacaaIXaaabaWaaO % aaaeaaciGGJbGaai4BaiaacohacaWG4baaleqaaaaaaaa!430B! y' = \frac{1}{{\sqrt {\sin x} }} - \frac{1}{{\sqrt {\cos x} }}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaigdaaeaadaGcaaqaaiGacohacaGGPbGa % aiOBaiaadIhaaSqabaaaaOGaey4kaSYaaSaaaeaacaaIXaaabaWaaO % aaaeaaciGGJbGaai4BaiaacohacaWG4baaleqaaaaaaaa!4300! y' = \frac{1}{{\sqrt {\sin x} }} + \frac{1}{{\sqrt {\cos x} }}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiGacogacaGGVbGaai4CaiaadIhaaeaadaGc % aaqaaiGacohacaGGPbGaaiOBaiaadIhaaSqabaaaaOGaeyOeI0YaaS % aaaeaaciGGZbGaaiyAaiaac6gacaWG4baabaWaaOaaaeaaciGGJbGa % ai4BaiaacohacaWG4baaleqaaaaaaaa!493A! y' = \frac{{\cos x}}{{\sqrt {\sin x} }} - \frac{{\sin x}}{{\sqrt {\cos x} }}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiGacogacaGGVbGaai4CaiaadIhaaeaadaGc % aaqaaiGacohacaGGPbGaaiOBaiaadIhaaSqabaaaaOGaey4kaSYaaS % aaaeaaciGGZbGaaiyAaiaac6gacaWG4baabaWaaOaaaeaaciGGJbGa % ai4BaiaacohacaWG4baaleqaaaaaaaa!492F! y' = \frac{{\cos x}}{{\sqrt {\sin x} }} + \frac{{\sin x}}{{\sqrt {\cos x} }}\)
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Câu 34:
Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpcaaI % YaGaci4CaiaacMgacaGGUbWaaOaaaeaacaWG4baaleqaaaaa!411B! y = f\left( x \right) = 2\sin \sqrt x \). Đạo hàm của hàm số y là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIYaGaci4yaiaac+gacaGGZbWaaOaaaeaacaWG4baa % leqaaaaa!3D4A! y' = 2\cos \sqrt x \)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaigdaaeaadaGcaaqaaiaadIhaaSqabaaa % aOGaci4yaiaac+gacaGGZbWaaOaaaeaacaWG4baaleqaaaaa!3E7B! y' = \frac{1}{{\sqrt x }}\cos \sqrt x \)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIYaWaaOaaaeaacaWG4baaleqaaOGaaiOlaiGacoga % caGGVbGaai4CamaalaaabaGaaGymaaqaamaakaaabaGaamiEaaWcbe % aaaaaaaa!3FE9! y' = 2\sqrt x .\cos \frac{1}{{\sqrt x }}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaigdaaeaadaGcaaqaaiaadIhaaSqabaGc % caGGUaGaci4yaiaac+gacaGGZbWaaOaaaeaacaWG4baaleqaaaaaaa % a!3F2D! y' = \frac{1}{{\sqrt x .\cos \sqrt x }}\)
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Câu 35:
Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacogacaGGVbGaaiiDamaaCaaaleqabaGaaGOmaaaakmaalaaa % baGaamiEaaqaaiaaisdaaaaaaa!3D8A! y = {\cot ^2}\frac{x}{4}\). Khi đó nghiệm của phương trình y’=0 là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiWdaNaey % 4kaSIaam4AaiaaikdacqaHapaCaaa!3BFC! \pi + k2\pi \)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiWdaNaey % 4kaSIaam4AaiaaikdacqaHapaCaaa!3BFC! 2\pi + k4\pi \)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiabec % 8aWjabgUcaRiaadUgacqaHapaCaaa!3BFC! 2\pi + k\pi \)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiabec % 8aWjabgUcaRiaadUgacqaHapaCaaa!3BFC! \pi + k\pi \)
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Câu 36:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaci4CaiaacMgacaGGUbGaamiEaiabgkHiTiaadIha % ciGGJbGaai4BaiaacohacaWG4baabaGaci4yaiaac+gacaGGZbGaam % iEaiabgUcaRiaadIhaciGGZbGaaiyAaiaac6gacaWG4baaaaaa!4B1B! y = \frac{{\sin x - x\cos x}}{{\cos x + x\sin x}}\) có đạo hàm bằng
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaWG4bWaaWbaaSqabeaacaaIYaaaaOGaaiOlaiGacohacaGG % PbGaaiOBaiaaikdacaWG4baabaGaaiikaiGacogacaGGVbGaai4Cai % aadIhacqGHRaWkcaWG4bGaci4CaiaacMgacaGGUbGaamiEaiaacMca % daahaaWcbeqaaiaaikdaaaaaaaaa!49EA! \frac{{ - {x^2}.\sin 2x}}{{{{(\cos x + x\sin x)}^2}}}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaWG4bWaaWbaaSqabeaacaaIYaaaaOGaaiOlaiGacohacaGG % PbGaaiOBamaaCaaaleqabaGaaGOmaaaakiaadIhaaeaacaGGOaGaci % 4yaiaac+gacaGGZbGaamiEaiabgUcaRiaadIhaciGGZbGaaiyAaiaa % c6gacaWG4bGaaiykamaaCaaaleqabaGaaGOmaaaaaaaaaa!4A21! \frac{{ - {x^2}.{{\sin }^2}x}}{{{{(\cos x + x\sin x)}^2}}}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaWG4bWaaWbaaSqabeaacaaIYaaaaOGaaiOlaiGacogacaGG % VbGaai4CaiaaikdacaWG4baabaGaaiikaiGacogacaGGVbGaai4Cai % aadIhacqGHRaWkcaWG4bGaci4CaiaacMgacaGGUbGaamiEaiaacMca % daahaaWcbeqaaiaaikdaaaaaaaaa!49E5! \frac{{ - {x^2}.\cos 2x}}{{{{(\cos x + x\sin x)}^2}}}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaada % WcaaqaaiaadIhaaeaaciGGJbGaai4BaiaacohacaWG4bGaey4kaSIa % amiEaiGacohacaGGPbGaaiOBaiaadIhaaaaacaGLOaGaayzkaaWaaW % baaSqabeaacaaIYaaaaaaa!42F7! {\left( {\frac{x}{{\cos x + x\sin x}}} \right)^2}\)
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Câu 37:
Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaci4yaiaac+gacaGGZbGaaGOmaiaadIhaaeaacaaI % ZaGaamiEaiabgUcaRiaaigdaaaaaaa!3FEA! y= \frac{{\cos 2x}}{{3x + 1}}\) là
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiabgkHiTiaaikdaciGGZbGaaiyAaiaac6ga % caaIYaGaamiEamaabmaabaGaaG4maiaadIhacqGHRaWkcaaIXaaaca % GLOaGaayzkaaGaeyOeI0IaaG4maiGacogacaGGVbGaai4Caiaaikda % caWG4baabaWaaeWaaeaacaaIZaGaamiEaiabgUcaRiaaigdaaiaawI % cacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaOGaaiOlaaaa!5087! y' = \frac{{ - 2\sin 2x\left( {3x + 1} \right) - 3\cos 2x}}{{{{\left( {3x + 1} \right)}^2}}}.\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiabgkHiTiaaikdaciGGZbGaaiyAaiaac6ga % caaIYaGaamiEamaabmaabaGaaG4maiaadIhacqGHRaWkcaaIXaaaca % GLOaGaayzkaaGaeyOeI0IaaG4maiGacogacaGGVbGaai4Caiaaikda % caWG4baabaGaaG4maiaadIhacqGHRaWkcaaIXaaaaiaac6caaaa!4E0B! y' = \frac{{ - 2\sin 2x\left( {3x + 1} \right) - 3\cos 2x}}{{3x + 1}}.\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiabgkHiTiGacohacaGGPbGaaiOBaiaaikda % caWG4bWaaeWaaeaacaaIZaGaamiEaiabgUcaRiaaigdaaiaawIcaca % GLPaaacqGHsislcaaIZaGaci4yaiaac+gacaGGZbGaaGOmaiaadIha % aeaadaqadaqaaiaaiodacaWG4bGaey4kaSIaaGymaaGaayjkaiaawM % caamaaCaaaleqabaGaaGOmaaaaaaGccaGGUaaaaa!4FCB! y' = \frac{{ - \sin 2x\left( {3x + 1} \right) - 3\cos 2x}}{{{{\left( {3x + 1} \right)}^2}}}.\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaikdaciGGZbGaaiyAaiaac6gacaaIYaGa % amiEamaabmaabaGaaG4maiaadIhacqGHRaWkcaaIXaaacaGLOaGaay % zkaaGaey4kaSIaaG4maiGacogacaGGVbGaai4CaiaaikdacaWG4baa % baWaaeWaaeaacaaIZaGaamiEaiabgUcaRiaaigdaaiaawIcacaGLPa % aadaahaaWcbeqaaiaaikdaaaaaaOGaaiOlaaaa!4F8F! y' = \frac{{2\sin 2x\left( {3x + 1} \right) + 3\cos 2x}}{{{{\left( {3x + 1} \right)}^2}}}.\)
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Câu 38:
Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaaGymaiabgUcaRiGacohacaGGPbGaaiOBaiaadIha % aeaacaaIXaGaey4kaSIaci4yaiaac+gacaGGZbGaamiEaaaaaaa!42E6! y = \frac{{1 + \sin x}}{{1 + \cos x}}\). Xét hai kết quả:
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0ZaaSaaaeaadaqadaqaaiGacogacaGGVbGaai4CaiaadIha % cqGHsislciGGZbGaaiyAaiaac6gacaWG4baacaGLOaGaayzkaaWaae % WaaeaacaaIXaGaey4kaSIaci4yaiaac+gacaGGZbGaamiEaiabgUca % RiGacohacaGGPbGaaiOBaiaadIhaaiaawIcacaGLPaaaaeaadaqada % qaaiaaigdacqGHRaWkciGGJbGaai4BaiaacohacaWG4baacaGLOaGa % ayzkaaWaaWbaaSqabeaacaaIYaaaaaaaaaa!55BA! (I) y' = \frac{{\left( {\cos x - \sin x} \right)\left( {1 + \cos x + \sin x} \right)}}{{{{\left( {1 + \cos x} \right)}^2}}}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0ZaaSaaaeaacaaIXaGaey4kaSIaci4yaiaac+gacaGGZbGa % amiEaiabgUcaRiGacohacaGGPbGaaiOBaiaadIhaaeaadaqadaqaai % aaigdacqGHRaWkciGGJbGaai4BaiaacohacaWG4baacaGLOaGaayzk % aaWaaWbaaSqabeaacaaIYaaaaaaaaaa!4A16! (II)y' = \frac{{1 + \cos x + \sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}\)
Kết quả nào đúng ?
A. Cả hai đều sai
B. Chỉ (I)
C. Chỉ (II)
D. Cả hai đều đúng
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Câu 39:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maabmaabaGaaGymaiabgUcaRiGacohacaGGPbGaaiOBaiaadIha % aiaawIcacaGLPaaadaqadaqaaiaaigdacqGHRaWkciGGJbGaai4Bai % aacohacaWG4baacaGLOaGaayzkaaaaaa!45E8! y = \left( {1 + \sin x} \right)\left( {1 + \cos x} \right)\) có đạo hàm là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qaceWG5bGbauaacqGH9aqpciGGJbGaai4BaiaacohacaWG4bGaeyOe % I0Iaci4CaiaacMgacaGGUbGaamiEaiabgUcaRiaaigdaaaa!4252! y' = \cos x - \sin x + 1\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0Jaci4yaiaac+gacaGGZbGaamiEaiabgUcaRiGacohacaGG % PbGaaiOBaiaadIhacqGHRaWkciGGJbGaai4BaiaacohacaaIYaGaam % iEaaaa!45F8! y' = \cos x + \sin x + \cos 2x\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qaceWG5bGbauaacqGH9aqpciGGJbGaai4BaiaacohacaWG4bGaeyOe % I0Iaci4CaiaacMgacaGGUbGaamiEaiabgUcaRiGacogacaGGVbGaai % 4CaiaaikdacaWG4baaaa!4623! y' = \cos x - \sin x + \cos 2x\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qaceWG5bGbauaacqGH9aqpciGGJbGaai4BaiaacohacaWG4bGaey4k % aSIaci4CaiaacMgacaGGUbGaamiEaiabgUcaRiaaigdaaaa!4247! y' = \cos x + \sin x + 1\)
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Câu 40:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadIhadaahaaWcbeqaaiaaikdaaaGccaGGUaGaci4yaiaac+ga % caGGZbGaamiEaaaa!3E6A! y = {x^2}.\cos x\) có đạo hàm là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIYaGaamiEaiaac6caciGGJbGaai4BaiaacohacaWG % 4bGaeyOeI0IaamiEamaaCaaaleqabaGaaGOmaaaakiGacohacaGGPb % GaaiOBaiaadIhaaaa!4590! y' = 2x.\cos x - {x^2}\sin x\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIYaGaamiEaiaac6caciGGJbGaai4BaiaacohacaWG % 4bGaey4kaSIaamiEamaaCaaaleqabaGaaGOmaaaakiGacohacaGGPb % GaaiOBaiaadIhaaaa!4585! y' = 2x.\cos x + {x^2}\sin x\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIYaGaamiEaiaac6cacaGGZbGaaiyAaiaac6gacaWG % 4bGaeyOeI0IaamiEamaaCaaaleqabaGaaGOmaaaakiGacogacaGGVb % Gaai4CaiaadIhaaaa!458E! y' = 2x.sinx - {x^2}\cos x\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIYaGaamiEaiaac6caciGGZbGaaiyAaiaac6gacaWG % 4bGaey4kaSIaamiEamaaCaaaleqabaGaaGOmaaaakiGacogacaGGVb % Gaai4CaiaadIhaaaa!4585! y' = 2x.\sin x + {x^2}\cos x\)
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Câu 41:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaci4CaiaacMgacaGGUbGaaiiEaaqaaiaadIhaaaaa % aa!3CD9! y = \frac{{{\mathop{\rm sinx}\nolimits} }}{x}\)có đạo hàm là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaadIhaciGGJbGaai4BaiaacohacaWG4bGa % ey4kaSIaci4CaiaacMgacaGGUbGaamiEaaqaaiaadIhadaahaaWcbe % qaaiaaikdaaaaaaaaa!441D! y' = \frac{{x\cos x + \sin x}}{{{x^2}}}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaadIhaciGGJbGaai4BaiaacohacaWG4bGa % eyOeI0Iaci4CaiaacMgacaGGUbGaamiEaaqaaiaadIhadaahaaWcbe % qaaiaaikdaaaaaaaaa!4428! y' = \frac{{x\cos x - \sin x}}{{{x^2}}}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaadIhaciGGZbGaaiyAaiaac6gacaWG4bGa % ey4kaSIaci4yaiaac+gacaGGZbGaamiEaaqaaiaadIhadaahaaWcbe % qaaiaaikdaaaaaaaaa!441D! y' = \frac{{x\sin x + \cos x}}{{{x^2}}}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaadIhaciGGZbGaaiyAaiaac6gacaWG4bGa % eyOeI0Iaci4yaiaac+gacaGGZbGaamiEaaqaaiaadIhadaahaaWcbe % qaaiaaikdaaaaaaaaa!4428! y' = \frac{{x\sin x - \cos x}}{{{x^2}}}\)
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Câu 42:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOmaaaakiaadIha % caGGUaGaci4yaiaac+gacaGGZbGaamiEaaaa!4142! y = {\sin ^2}x.\cos x\)có đạo hàm là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpciGGZbGaaiyAaiaac6gacaGG4bWaaeWaaeaacaaIZaGa % ci4yaiaac+gacaGGZbWaaWbaaSqabeaacaaIYaaaaOGaamiEaiabgk % HiTiaaigdaaiaawIcacaGLPaaaaaa!4527! y' = {\mathop{\rm sinx}\nolimits} \left( {3{{\cos }^2}x - 1} \right)\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpciGGZbGaaiyAaiaac6gacaGG4bWaaeWaaeaacaaIZaGa % ci4yaiaac+gacaGGZbWaaWbaaSqabeaacaaIYaaaaOGaamiEaiabgU % caRiaaigdaaiaawIcacaGLPaaaaaa!451C! y' = {\mathop{\rm sinx}\nolimits} \left( {3{{\cos }^2}x + 1} \right)\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpciGGZbGaaiyAaiaac6gacaGG4bWaaeWaaeaaciGGJbGa % ai4BaiaacohadaahaaWcbeqaaiaaikdaaaGccaWG4bGaey4kaSIaaG % ymaaGaayjkaiaawMcaaaaa!445F! y' = {\mathop{\rm sinx}\nolimits} \left( {{{\cos }^2}x + 1} \right)\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpciGGZbGaaiyAaiaac6gacaGG4bWaaeWaaeaaciGGJbGa % ai4BaiaacohadaahaaWcbeqaaiaaikdaaaGccaWG4bGaeyOeI0IaaG % ymaaGaayjkaiaawMcaaaaa!446A! y' = {\mathop{\rm sinx}\nolimits} \left( {{{\cos }^2}x - 1} \right)\)
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Câu 43:
Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacogacaGGVbGaai4CamaabmaabaGaciiDaiaacggacaGGUbGa % amiEaaGaayjkaiaawMcaaaaa!4022! y = \cos \left( {\tan x} \right)\) bằng
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4CaiaacM % gacaGGUbWaaeWaaeaaciGG0bGaaiyyaiaac6gacaWG4baacaGLOaGa % ayzkaaGaeyyXIC9aaSaaaeaacaaIXaaabaGaci4yaiaac+gacaGGZb % WaaWbaaSqabeaacaaIYaaaaOGaamiEaaaacqGHflY1aaa!4845! \sin \left( {\tan x} \right) \cdot \frac{1}{{{{\cos }^2}x}} \cdot \)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Iaci % 4CaiaacMgacaGGUbWaaeWaaeaaciGG0bGaaiyyaiaac6gacaWG4baa % caGLOaGaayzkaaGaeyyXIC9aaSaaaeaacaaIXaaabaGaci4yaiaac+ % gacaGGZbWaaWbaaSqabeaacaaIYaaaaOGaamiEaaaacqGHflY1aaa!4932! - \sin \left( {\tan x} \right) \cdot \frac{1}{{{{\cos }^2}x}} \cdot \)
C. \(sin(tanx)\)
D. \(-sin (tan x)\)
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Câu 44:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iabgkHiTmaalaaabaGaaGymaaqaaiaaikdaaaGaci4CaiaacMga % caGGUbWaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaiodaaaGaeyOeI0 % IaamiEamaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaaaaa!4434! y = - \frac{1}{2}\sin \left( {\frac{\pi }{3} - {x^2}} \right)\) có đạo hàm là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiaac6 % caciGGJbGaai4BaiaacohadaqadaqaamaalaaabaGaeqiWdahabaGa % aG4maaaacqGHsislcaWG4bWaaWbaaSqabeaacaaIYaaaaaGccaGLOa % Gaayzkaaaaaa!4166! x.\cos \left( {\frac{\pi }{3} - {x^2}} \right)\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaaGOmaaaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaci4y % aiaac+gacaGGZbWaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaiodaaa % GaeyOeI0IaamiEaaGaayjkaiaawMcaaaaa!423B! \frac{1}{2}{x^2}\cos \left( {\frac{\pi }{3} - x} \right)\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaaGOmaaaacaWG4bGaci4CaiaacMgacaGGUbWaaeWaaeaa % daWcaaqaaiabec8aWbqaaiaaiodaaaGaeyOeI0IaamiEaaGaayjkai % aawMcaaaaa!414D! \frac{1}{2}x\sin \left( {\frac{\pi }{3} - x} \right)\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaaGOmaaaacaWG4bGaci4yaiaac+gacaGGZbWaaeWaaeaa % daWcaaqaaiabec8aWbqaaiaaiodaaaGaeyOeI0IaamiEamaaCaaale % qabaGaaGOmaaaaaOGaayjkaiaawMcaaaaa!423B! \frac{1}{2}x\cos \left( {\frac{\pi }{3} - {x^2}} \right)\)
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Câu 45:
Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maakaaabaGaci4Caiaa % cMgacaGGUbGaaG4maiaadIhaaSqabaaaaa!3F17! f\left( x \right) = \sqrt {\sin 3x} \) là
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIZaGaci4yaiaac+gacaGGZbGaaG4maiaadIhaaeaadaGcaaqaaiGa % cohacaGGPbGaaiOBaiaaiodacaWG4baaleqaaaaakiabgwSixdaa!424E! \frac{{3\cos 3x}}{{\sqrt {\sin 3x} }} \cdot \)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIZaGaci4yaiaac+gacaGGZbGaaG4maiaadIhaaeaacaaIYaWaaOaa % aeaaciGGZbGaaiyAaiaac6gacaaIZaGaamiEaaWcbeaaaaGccqGHfl % Y1aaa!430A! \frac{{3\cos 3x}}{{2\sqrt {\sin 3x} }} \cdot \)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0YaaS % aaaeaacaaIZaGaci4yaiaac+gacaGGZbGaaG4maiaadIhaaeaacaaI % YaWaaOaaaeaaciGGZbGaaiyAaiaac6gacaaIZaGaamiEaaWcbeaaaa % GccqGHflY1aaa!43F7! - \frac{{3\cos 3x}}{{2\sqrt {\sin 3x} }} \cdot \)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaci % GGJbGaai4BaiaacohacaaIZaGaamiEaaqaaiaaikdadaGcaaqaaiGa % cohacaGGPbGaaiOBaiaaiodacaWG4baaleqaaaaakiabgwSixdaa!424D! \frac{{\cos 3x}}{{2\sqrt {\sin 3x} }} \cdot \)
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Câu 46:
Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaikdaciGGZbGaaiyA % aiaac6gacaaIYaGaamiEaiabgUcaRiGacogacaGGVbGaai4Caiaaik % dacaWG4baaaa!4525! f\left( x \right) = 2\sin 2x + \cos 2x\) là
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGinaiGaco % gacaGGVbGaai4CaiaaikdacaWG4bGaey4kaSIaaGOmaiGacohacaGG % PbGaaiOBaiaaikdacaWG4baaaa!416C! 4\cos 2x + 2\sin 2x\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiGaco % gacaGGVbGaai4CaiaaikdacaWG4bGaeyOeI0IaaGOmaiGacohacaGG % PbGaaiOBaiaaikdacaWG4baaaa!4175! 2\cos 2x - 2\sin 2x\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGinaiGaco % gacaGGVbGaai4CaiaaikdacaWG4bGaey4kaSIaaGOmaiGacohacaGG % PbGaaiOBaiaaikdacaWG4baaaa!416C! 4\cos 2x - 2\sin 2x\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % inaiGacogacaGGVbGaai4CaiaaikdacaWG4bGaeyOeI0IaaGOmaiGa % cohacaGGPbGaaiOBaiaaikdacaWG4baaaa!4264! - 4\cos 2x - 2\sin 2x\)
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Câu 47:
Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacohacaGGPbGaaiOBamaabmaabaWaaSaaaeaacqaHapaCaeaa % caaIYaaaaiabgkHiTiaaikdacaWG4baacaGLOaGaayzkaaaaaa!4187! y = \sin \left( {\frac{\pi }{2} - 2x} \right)\) là y’ bằng
A. -2sin2x
B. 2sin2x
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Iaci % 4yaiaac+gacaGGZbWaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaikda % aaGaeyOeI0IaaGOmaiaadIhaaiaawIcacaGLPaaaaaa!406B! - \cos \left( {\frac{\pi }{2} - 2x} \right)\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0Iaci % 4yaiaac+gacaGGZbWaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaikda % aaGaeyOeI0IaaGOmaiaadIhaaiaawIcacaGLPaaaaaa!406B! \cos \left( {\frac{\pi }{2} - 2x} \right)\)
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Câu 48:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacshacaGGHbGaaiOBaiaaiEdacaWG4baaaa!3C86! y = \tan 7x\) có đạo hàm là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI3aaabaGaci4yaiaac+gacaGGZbWaaWbaaSqabeaacaaIYaaaaOGa % aG4naiaadIhaaaaaaa!3C48! \frac{7}{{{{\cos }^2}7x}}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0YaaS % aaaeaacaaI3aaabaGaci4yaiaac+gacaGGZbWaaWbaaSqabeaacaaI % YaaaaOGaaG4naiaadIhaaaaaaa!3D35! - \frac{7}{{{{\cos }^2}7x}}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0YaaS % aaaeaacaaI3aaabaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaI % YaaaaOGaaG4naiaadIhaaaaaaa!3D3A! - \frac{7}{{{{\sin }^2}7x}}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI3aGaamiEaaqaaiGacogacaGGVbGaai4CamaaCaaaleqabaGaaGOm % aaaakiaaiEdacaWG4baaaaaa!3D45! \frac{{7x}}{{{{\cos }^2}7x}}\)
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Câu 49:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaaGymaaqaaiaaikdaaaWaaeWaaeaacaaIXaGaey4k % aSIaciiDaiaacggacaGGUbGaamiEaaGaayjkaiaawMcaamaaCaaale % qabaGaaGOmaaaaaaa!415B! y = \frac{1}{2}{\left( {1 + \tan x} \right)^2}\) có đạo hàm
A. \(y' = 1 + tanx\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaqadaqaaiaaigdacqGHRaWkciGG0bGaaiyyaiaac6ga % caWG4baacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaaa!407F! y' = {\left( {1 + \tan x} \right)^2}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaqadaqaaiaaigdacqGHRaWkciGG0bGaaiyyaiaac6ga % caWG4baacaGLOaGaayzkaaWaaeWaaeaacaaIXaGaey4kaSIaciiDai % aacggacaGGUbWaaWbaaSqabeaacaaIYaaaaOGaamiEaaGaayjkaiaa % wMcaaaaa!477D! y' = \left( {1 + \tan x} \right)\left( {1 + {{\tan }^2}x} \right)\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIXaGaey4kaSIaciiDaiaacggacaGGUbWaaWbaaSqa % beaacaaIYaaaaOGaamiEaaaa!3F00! y' = 1 + {\tan ^2}x\)
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Câu 50:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaaGymaaqaaiaaikdaaaGaci4yaiaac+gacaGG0bGa % amiEamaaCaaaleqabaGaaGOmaaaaaaa!3E39! y = \frac{1}{2}\cot {x^2}\) có đạo hàm là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaWG4baabaGaaGOmaiGacohacaGGPbGaaiOBaiaadIhadaah % aaWcbeqaaiaaikdaaaaaaOGaeyyXICnaaa!3FBC! \frac{{ - x}}{{2\sin {x^2}}} \cdot \)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4baabaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaIYaaaaOGa % amiEamaaCaaaleqabaGaaGOmaaaaaaGccqGHflY1aaa!3F06! \frac{x}{{{{\sin }^2}{x^2}}} \cdot \)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaWG4baabaGaci4CaiaacMgacaGGUbGaamiEamaaCaaaleqa % baGaaGOmaaaaaaGccqGHflY1aaa!3F00! \frac{{ - x}}{{\sin {x^2}}} \cdot \)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaWG4baabaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaI % YaaaaOGaamiEamaaCaaaleqabaGaaGOmaaaaaaGccqGHflY1aaa!3FF3! \frac{{ - x}}{{{{\sin }^2}{x^2}}} \cdot \)