Tính đạo hàm của hàm số sau \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacohacaGGPbGaaiOBamaakaaabaGaaGOmaiabgUcaRiaadIha % daahaaWcbeqaaiaaikdaaaaabeaaaaa!3E64! y = \sin \sqrt {2 + {x^2}} \)
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpciGGJbGaai4BaiaacohadaGcaaqaaiaaikdacqGHRaWk % caWG4bWaaWbaaSqabeaacaaIYaaaaaqabaGccaGGUaWaaeWaaeaada % GcaaqaaiaaikdacqGHRaWkcaWG4bWaaWbaaSqabeaacaaIYaaaaaqa % baaakiaawIcacaGLPaaadaahaaWcbeqaaiaac+caaaGccqGH9aqpci % GGJbGaai4BaiaacohadaGcaaqaaiaaikdacqGHRaWkcaWG4bWaaWba % aSqabeaacaaIYaaaaaqabaGccaGGUaWaaSaaaeaadaqadaqaaiaaik % dacqGHRaWkcaWG4bWaaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzk % aaWaaWbaaSqabeaacaGGVaaaaaGcbaGaaGOmamaakaaabaGaaGOmai % abgUcaRiaadIhadaahaaWcbeqaaiaaikdaaaaabeaaaaGccqGH9aqp % daWcaaqaaiaadIhaaeaadaGcaaqaaiaaikdacqGHRaWkcaWG4bWaaW % baaSqabeaacaaIYaaaaaqabaaaaOGaaiOlaiGacogacaGGVbGaai4C % amaakaaabaGaaGOmaiabgUcaRiaadIhadaahaaWcbeqaaiaaikdaaa % aabeaakiaac6caaaa!65F1! y' = \cos \sqrt {2 + {x^2}} .{\left( {\sqrt {2 + {x^2}} } \right)^/} = \cos \sqrt {2 + {x^2}} .\frac{{{{\left( {2 + {x^2}} \right)}^/}}}{{2\sqrt {2 + {x^2}} }} = \frac{x}{{\sqrt {2 + {x^2}} }}.\cos \sqrt {2 + {x^2}} .\)