Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaci4yaiaac+gacaGGZbGaaGOmaiaadIhaaeaacaaI % XaGaeyOeI0Iaci4CaiaacMgacaGGUbGaamiEaaaaaaa!4211! y = \frac{{\cos 2x}}{{1 - \sin x}}\). Tính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cadaqadaqaamaalaaabaGaeqiWdahabaGaaGOnaaaaaiaawIcacaGL % Paaaaaa!3BB3! y'\left( {\frac{\pi }{6}} \right)\) bằng:
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaamaabmaabaGaci4yaiaac+gacaGGZbGaaGOm % aiaadIhaaiaawIcacaGLPaaacaGGNaGaaiOlamaabmaabaGaaGymai % abgkHiTiGacohacaGGPbGaaiOBaiaadIhaaiaawIcacaGLPaaacqGH % sislciGGJbGaai4BaiaacohacaaIYaGaamiEamaabmaabaGaaGymai % abgkHiTiGacohacaGGPbGaaiOBaiaadIhaaiaawIcacaGLPaaacaGG % NaaabaWaaeWaaeaacaaIXaGaeyOeI0Iaci4CaiaacMgacaGGUbGaam % iEaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaaGccqGH9aqp % daWcaaqaaiabgkHiTiaaikdaciGGZbGaaiyAaiaac6gacaaIYaGaam % iEamaabmaabaGaaGymaiabgkHiTiGacohacaGGPbGaaiOBaiaadIha % aiaawIcacaGLPaaacqGHRaWkciGGJbGaai4BaiaacohacaaIYaGaam % iEaiaac6cacaGGJbGaai4BaiaacohacaGG4baabaWaaeWaaeaacaaI % XaGaeyOeI0Iaci4CaiaacMgacaGGUbGaamiEaaGaayjkaiaawMcaam % aaCaaaleqabaGaaGOmaaaaaaaaaa!7C80! y' = \frac{{\left( {\cos 2x} \right)'.\left( {1 - \sin x} \right) - \cos 2x\left( {1 - \sin x} \right)'}}{{{{\left( {1 - \sin x} \right)}^2}}} = \frac{{ - 2\sin 2x\left( {1 - \sin x} \right) + \cos 2x.cosx}}{{{{\left( {1 - \sin x} \right)}^2}}}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cadaqadaqaamaalaaabaGaeqiWdahabaGaaGOnaaaaaiaawIcacaGL % PaaacqGH9aqpdaWcaaqaaiabgkHiTiaaikdacaGGUaWaaSaaaeaada % GcaaqaaiaaiodaaSqabaaakeaacaaIYaaaamaabmaabaGaaGymaiab % gkHiTmaalaaabaGaaGymaaqaaiaaikdaaaaacaGLOaGaayzkaaGaey % 4kaSYaaSaaaeaacaaIXaaabaGaaGOmaaaacaGGUaWaaSaaaeaadaGc % aaqaaiaaiodaaSqabaaakeaacaaIYaaaaaqaamaabmaabaGaaGymai % abgkHiTmaalaaabaGaaGymaaqaaiaaikdaaaaacaGLOaGaayzkaaWa % aWbaaSqabeaacaaIYaaaaaaakiabg2da9maalaaabaGaeyOeI0YaaS % aaaeaadaGcaaqaaiaaiodaaSqabaaakeaacaaIYaaaaiabgUcaRmaa % laaabaWaaOaaaeaacaaIZaaaleqaaaGcbaGaaGinaaaaaeaadaWcaa % qaaiaaigdaaeaacaaI0aaaaaaacqGH9aqpcaaI0aWaaeWaaeaacqGH % sisldaWcaaqaamaakaaabaGaaG4maaWcbeaaaOqaaiaaikdaaaGaey % 4kaSYaaSaaaeaadaGcaaqaaiaaiodaaSqabaaakeaacaaI0aaaaaGa % ayjkaiaawMcaaiabg2da9iabgkHiTiaaikdadaGcaaqaaiaaiodaaS % qabaGccqGHRaWkdaGcaaqaaiaaiodaaSqabaGccqGH9aqpcqGHsisl % daGcaaqaaiaaiodaaSqabaaaaa!6863! y'\left( {\frac{\pi }{6}} \right) = \frac{{ - 2.\frac{{\sqrt 3 }}{2}\left( {1 - \frac{1}{2}} \right) + \frac{1}{2}.\frac{{\sqrt 3 }}{2}}}{{{{\left( {1 - \frac{1}{2}} \right)}^2}}} = \frac{{ - \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{4}}}{{\frac{1}{4}}} = 4\left( { - \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{4}} \right) = - 2\sqrt 3 + \sqrt 3 = - \sqrt 3 \)
