Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpdaGc % aaqaaiGacshacaGGHbGaaiOBaiaadIhacqGHRaWkciGGJbGaai4Bai % aacshacaWG4baaleqaaaaa!450B! y = f\left( x \right) = \sqrt {\tan x + \cot x} \). Giá trị \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cadaqadaqaamaalaaabaGaeqiWdahabaGaaGinaaaaaiaawIcacaGL % Paaaaaa!3B9E! f'\left( {\frac{\pi }{4}} \right)\) bằng:
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maakaaabaGaciiDaiaacggacaGGUbGaamiEaiabgUcaRiGacoga % caGGVbGaaiiDaiaadIhaaSqabaGccqGHshI3caWG5bWaaWbaaSqabe % aacaaIYaaaaOGaeyypa0JaciiDaiaacggacaGGUbGaamiEaiabgUca % RiGacogacaGGVbGaaiiDaiaadIhacqGHshI3caWG5bGaai4jaiaac6 % cacaaIYaGaamyEaiabg2da9maalaaabaGaaGymaaqaaiGacogacaGG % VbGaai4CamaaCaaaleqabaGaaGOmaaaakiaadIhaaaGaeyOeI0YaaS % aaaeaacaaIXaaabaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaI % YaaaaOGaamiEaaaaaaa!61F9! y = \sqrt {\tan x + \cot x} \Rightarrow {y^2} = \tan x + \cot x \Rightarrow y'.2y = \frac{1}{{{{\cos }^2}x}} - \frac{1}{{{{\sin }^2}x}}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yEaiaacEcacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaWaaOaaaeaa % ciGG0bGaaiyyaiaac6gacaWG4bGaey4kaSIaci4yaiaac+gacaGG0b % GaamiEaaWcbeaaaaGcdaqadaqaamaalaaabaGaaGymaaqaaiGacoga % caGGVbGaai4CamaaCaaaleqabaGaaGOmaaaakiaadIhaaaGaeyOeI0 % YaaSaaaeaacaaIXaaabaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaa % caaIYaaaaOGaamiEaaaaaiaawIcacaGLPaaaaaa!52C4! \Rightarrow y' = \frac{1}{{2\sqrt {\tan x + \cot x} }}\left( {\frac{1}{{{{\cos }^2}x}} - \frac{1}{{{{\sin }^2}x}}} \right)\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cadaqadaqaamaalaaabaGaeqiWdahabaGaaGinaaaaaiaawIcacaGL % PaaacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaWaaOaaaeaaciGG0b % Gaaiyyaiaac6gadaWcaaqaaiabec8aWbqaaiaaisdaaaGaey4kaSIa % ci4yaiaac+gacaGG0bWaaSaaaeaacqaHapaCaeaacaaI0aaaaaWcbe % aaaaGcdaqadaqaamaalaaabaGaaGymaaqaaiGacogacaGGVbGaai4C % amaaCaaaleqabaGaaGOmaaaakmaabmaabaWaaSaaaeaacqaHapaCae % aacaaI0aaaaaGaayjkaiaawMcaaaaacqGHsisldaWcaaqaaiaaigda % aeaaciGGZbGaaiyAaiaac6gadaahaaWcbeqaaiaaikdaaaGcdaqada % qaamaalaaabaGaeqiWdahabaGaaGinaaaaaiaawIcacaGLPaaaaaaa % caGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmamaaka % aabaGaaGOmaaWcbeaaaaGcdaqadaqaaiaaikdacqGHsislcaaIYaaa % caGLOaGaayzkaaGaeyypa0JaaGimaaaa!66CE! $'\left( {\frac{\pi }{4}} \right) = \frac{1}{{2\sqrt {\tan \frac{\pi }{4} + \cot \frac{\pi }{4}} }}\left( {\frac{1}{{{{\cos }^2}\left( {\frac{\pi }{4}} \right)}} - \frac{1}{{{{\sin }^2}\left( {\frac{\pi }{4}} \right)}}} \right) = \frac{1}{{2\sqrt 2 }}\left( {2 - 2} \right) = 0\)