Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpdaWc % aaqaaiaaigdaaeaadaGcaaqaaiGacohacaGGPbGaaiOBaiaadIhaaS % qabaaaaaaa!412A! y = f\left( x \right) = \frac{1}{{\sqrt {\sin x} }}\). Giá trị \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cadaqadaqaamaalaaabaGaeqiWdahabaGaaGOmaaaaaiaawIcacaGL % Paaaaaa!3B9C! f'\left( {\frac{\pi }{2}} \right)\) bằng:
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaaGymaaqaamaakaaabaGaci4CaiaacMgacaGGUbGa % amiEaaWcbeaaaaGccqGHshI3caWG5bWaaWbaaSqabeaacaaIYaaaaO % Gaeyypa0ZaaSaaaeaacaaIXaaabaGaci4CaiaacMgacaGGUbGaamiE % aaaacqGHshI3caWG5bGaai4jaiaaikdacaWG5bGaeyypa0ZaaSaaae % aacqGHsislciGGJbGaai4BaiaacohacaWG4baabaGaci4CaiaacMga % caGGUbWaaWbaaSqabeaacaaIYaaaaOGaamiEaaaaaaa!570C! y = \frac{1}{{\sqrt {\sin x} }} \Rightarrow {y^2} = \frac{1}{{\sin x}} \Rightarrow y'2y = \frac{{ - \cos x}}{{{{\sin }^2}x}}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yEaiaacEcacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaGaamyEaaaa % caGGUaWaaeWaaeaadaWcaaqaaiabgkHiTiGacogacaGGVbGaai4Cai % aadIhaaeaaciGGZbGaaiyAaiaac6gadaahaaWcbeqaaiaaikdaaaGc % caWG4baaaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaGymaaqaam % aalaaabaGaaGOmaaqaamaakaaabaGaci4CaiaacMgacaGGUbGaamiE % aaWcbeaaaaaaaOWaaeWaaeaadaWcaaqaaiabgkHiTiGacogacaGGVb % Gaai4CaiaadIhaaeaaciGGZbGaaiyAaiaac6gadaahaaWcbeqaaiaa % ikdaaaGccaWG4baaaaGaayjkaiaawMcaaiabg2da9maalaaabaGaey % OeI0YaaOaaaeaaciGGZbGaaiyAaiaac6gacaWG4baaleqaaaGcbaGa % aGOmaaaacaGGUaWaaSaaaeaaciGGJbGaai4BaiaacohacaWG4baaba % Gaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaIYaaaaOGaamiEaaaa % aaa!6B1C! \Rightarrow y' = \frac{1}{{2y}}.\left( {\frac{{ - \cos x}}{{{{\sin }^2}x}}} \right) = \frac{1}{{\frac{2}{{\sqrt {\sin x} }}}}\left( {\frac{{ - \cos x}}{{{{\sin }^2}x}}} \right) = \frac{{ - \sqrt {\sin x} }}{2}.\frac{{\cos x}}{{{{\sin }^2}x}}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cadaqadaqaamaalaaabaGaeqiWdahabaGaaGOmaaaaaiaawIcacaGL % PaaacqGH9aqpdaWcaaqaaiabgkHiTmaakaaabaGaci4CaiaacMgaca % GGUbWaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaikdaaaaacaGLOaGa % ayzkaaaaleqaaaGcbaGaaGOmaaaacaGGUaWaaSaaaeaaciGGJbGaai % 4BaiaacohadaqadaqaamaalaaabaGaeqiWdahabaGaaGOmaaaaaiaa % wIcacaGLPaaaaeaaciGGZbGaaiyAaiaac6gadaahaaWcbeqaaiaaik % daaaGcdaqadaqaamaalaaabaGaeqiWdahabaGaaGOmaaaaaiaawIca % caGLPaaaaaGaeyypa0ZaaSaaaeaacqGHsislcaaIXaaabaGaaGOmaa % aacaGGUaWaaSaaaeaacaaIWaaabaGaaGymaaaacqGH9aqpcaaIWaaa % aa!5C5F! f'\left( {\frac{\pi }{2}} \right) = \frac{{ - \sqrt {\sin \left( {\frac{\pi }{2}} \right)} }}{2}.\frac{{\cos \left( {\frac{\pi }{2}} \right)}}{{{{\sin }^2}\left( {\frac{\pi }{2}} \right)}} = \frac{{ - 1}}{2}.\frac{0}{1} = 0\)
