Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadIhadaahaaWcbeqaaiaaikdaaaGcciGG0bGaaiyyaiaac6ga % caWG4bGaey4kaSYaaOaaaeaacaWG4baaleqaaaaa!3FAF! y = {x^2}\tan x + \sqrt x \) là
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0ZaaeWaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaaGccaGL % OaGaayzkaaWaaWbaaSqabeaakiadacUHYaIOaaGaaeiDaiaabggaca % qGUbGaamiEaiaabUcadaqadaqaaiaabshacaqGHbGaaeOBaiaadIha % aiaawIcacaGLPaaadaahaaWcbeqaaOGamai4gkdiIcaacaGGUaGaam % iEamaaCaaaleqabaGaaGOmaaaakiabgUcaRmaabmaabaWaaOaaaeaa % caWG4baaleqaaaGccaGLOaGaayzkaaWaaWbaaSqabeaakiadacUHYa % IOaaGaeyO0H4TaamyEaiaacEcacqGH9aqpcaaIYaGaamiEaiGacsha % caGGHbGaaiOBaiaadIhacqGHRaWkdaWcaaqaaiaadIhadaahaaWcbe % qaaiaaikdaaaaakeaaciGGJbGaai4BaiaacohadaahaaWcbeqaaiaa % ikdaaaGccaWG4baaaiabgUcaRmaalaaabaGaaGymaaqaaiaaikdada % GcaaqaaiaadIhaaSqabaaaaOGaaiOlaaaa!6B47! y' = {\left( {{x^2}} \right)^\prime }{\rm{tan}}x{\rm{ + }}{\left( {{\rm{tan}}x} \right)^\prime }.{x^2} + {\left( {\sqrt x } \right)^\prime } \Rightarrow y' = 2x\tan x + \frac{{{x^2}}}{{{{\cos }^2}x}} + \frac{1}{{2\sqrt x }}.\)