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
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0ZaaSaaaeaadaWadaqaaiaaikdacqGHRaWkciGG0bGaaiyy % aiaac6gadaqadaqaaiaadIhacqGHRaWkdaWcaaqaaiaaigdaaeaaca % WG4baaaaGaayjkaiaawMcaaaGaay5waiaaw2faamaaCaaaleqabaGc % cWaGGBOmGikaaaqaaiaaikdadaGcaaqaaiaaikdacqGHRaWkciGG0b % Gaaiyyaiaac6gadaqadaqaaiaadIhacqGHRaWkdaWcaaqaaiaaigda % aeaacaWG4baaaaGaayjkaiaawMcaaaWcbeaaaaGccqGH9aqpdaWcaa % qaaiaaigdacqGHRaWkciGG0bGaaiyyaiaac6gadaahaaWcbeqaaiaa % ikdaaaGcdaqadaqaaiaadIhacqGHRaWkdaWcaaqaaiaaigdaaeaaca % WG4baaaaGaayjkaiaawMcaaaqaaiaaikdadaGcaaqaaiaaikdacqGH % RaWkciGG0bGaaiyyaiaac6gadaqadaqaaiaadIhacqGHRaWkdaWcaa % qaaiaaigdaaeaacaWG4baaaaGaayjkaiaawMcaaaWcbeaaaaGccqGH % flY1daqadaqaaiaadIhacqGHRaWkdaWcaaqaaiaaigdaaeaacaWG4b % aaaaGaayjkaiaawMcaamaaCaaaleqabaGccWaGGBOmGikaaiabg2da % 9maalaaabaGaaGymaiabgUcaRiGacshacaGGHbGaaiOBamaaCaaale % qabaGaaGOmaaaakmaabmaabaGaamiEaiabgUcaRmaalaaabaGaaGym % aaqaaiaadIhaaaaacaGLOaGaayzkaaaabaGaaGOmamaakaaabaGaaG % OmaiabgUcaRiGacshacaGGHbGaaiOBamaabmaabaGaamiEaiabgUca % RmaalaaabaGaaGymaaqaaiaadIhaaaaacaGLOaGaayzkaaaaleqaaa % aakiabgwSixpaabmaabaGaaGymaiabgkHiTmaalaaabaGaaGymaaqa % aiaadIhadaahaaWcbeqaaiaaikdaaaaaaaGccaGLOaGaayzkaaaaaa!9060! y' = \frac{{{{\left[ {2 + \tan \left( {x + \frac{1}{x}} \right)} \right]}^\prime }}}{{2\sqrt {2 + \tan \left( {x + \frac{1}{x}} \right)} }} = \frac{{1 + {{\tan }^2}\left( {x + \frac{1}{x}} \right)}}{{2\sqrt {2 + \tan \left( {x + \frac{1}{x}} \right)} }} \cdot {\left( {x + \frac{1}{x}} \right)^\prime } = \frac{{1 + {{\tan }^2}\left( {x + \frac{1}{x}} \right)}}{{2\sqrt {2 + \tan \left( {x + \frac{1}{x}} \right)} }} \cdot \left( {1 - \frac{1}{{{x^2}}}} \right)\)