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
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGOmaiGacogacaGGVbGaaiiDamaabmaabaGaci4yaiaa % c+gacaGGZbGaamiEaaGaayjkaiaawMcaaiaac6cadaqadaqaaiGaco % gacaGGVbGaaiiDamaabmaabaGaci4yaiaac+gacaGGZbGaamiEaaGa % ayjkaiaawMcaaaGaayjkaiaawMcaamaaCaaaleqabaGccWaGGBOmGi % kaaiabgUcaRmaalaaabaWaaeWaaeaaciGGZbGaaiyAaiaac6gacaWG % 4bGaaeylamaalaaabaGaeqiWdahabaGaaGOmaaaaaiaawIcacaGLPa % aadaahaaWcbeqaaOGamai4gkdiIcaaaeaacaaIYaWaaOaaaeaacaWG % ZbGaaeyAaiaab6gacaWG4bGaeyOeI0YaaSaaaeaacqaHapaCaeaaca % aIYaaaaaWcbeaaaaGccqGH9aqpcaaIYaGaci4yaiaac+gacaGG0bWa % aeWaaeaaciGGJbGaai4BaiaacohacaWG4baacaGLOaGaayzkaaWaaS % aaaeaacaaIXaaabaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaI % YaaaaOWaaeWaaeaaciGGJbGaai4BaiaacohacaWG4baacaGLOaGaay % zkaaaaaiaac6caciGGZbGaaiyAaiaac6gacaWG4bGaey4kaSYaaSaa % aeaaciGGJbGaai4BaiaacohacaWG4baabaGaaGOmamaakaaabaGaci % 4CaiaacMgacaGGUbGaamiEaiabgkHiTmaalaaabaGaeqiWdahabaGa % aGOmaaaaaSqabaaaaaaa!888E! y' = 2\cot \left( {\cos x} \right).{\left( {\cot \left( {\cos x} \right)} \right)^\prime } + \frac{{{{\left( {\sin x{\rm{ - }}\frac{\pi }{2}} \right)}^\prime }}}{{2\sqrt {s{\rm{in}}x - \frac{\pi }{2}} }} = 2\cot \left( {\cos x} \right)\frac{1}{{{{\sin }^2}\left( {\cos x} \right)}}.\sin x + \frac{{\cos x}}{{2\sqrt {\sin x - \frac{\pi }{2}} }}\)