Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgacaGGOaGaamiEaiaacMcacqGH9aqpdaGcaaqaaiGacsha % caGGHbGaaiOBaiaadIhacqGHRaWkciGGJbGaai4BaiaacshacaWG4b % aaleqaaaaa!44DA! y = f(x) = \sqrt {\tan x + \cot x} \). Giá trị \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaisdaaaaacaGLOaGaayzk % aaaaaa!3AFE! f'\left( {\frac{\pi }{4}} \right)\) bằng
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaadaqa % daqaaiaabshacaqGHbGaaeOBaiaadIhacqGHRaWkciGGJbGaai4Bai % aacshacaWG4baacaGLOaGaayzkaaWaaWbaaSqabeaakiadacUHYaIO % aaaabaGaaGOmamaakaaabaGaaeiDaiaabggacaqGUbGaamiEaiabgU % caRiGacogacaGGVbGaaiiDaiaadIhaaSqabaaaaOGaeyypa0ZaaSaa % aeaadaWcaaqaaiaaigdaaeaaciGGJbGaai4BaiaacohadaahaaWcbe % qaaiaaikdaaaGccaWG4baaaiabgkHiTmaalaaabaGaaGymaaqaaiGa % cohacaGGPbGaaiOBamaaCaaaleqabaGaaGOmaaaakiaadIhaaaaaba % GaaGOmamaakaaabaGaaeiDaiaabggacaqGUbGaamiEaiabgUcaRiGa % cogacaGGVbGaaiiDaiaadIhaaSqabaaaaOGaeyO0H4TabmOzayaafa % WaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaisdaaaaacaGLOaGaayzk % aaGaeyypa0JaaGimaiaac6caaaa!7163! f'\left( x \right) = \frac{{{{\left( {{\rm{tan}}x + \cot x} \right)}^\prime }}}{{2\sqrt {{\rm{tan}}x + \cot x} }} = \frac{{\frac{1}{{{{\cos }^2}x}} - \frac{1}{{{{\sin }^2}x}}}}{{2\sqrt {{\rm{tan}}x + \cot x} }} \Rightarrow f'\left( {\frac{\pi }{4}} \right) = 0.\)