Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaamiEaiab % gUcaRiaaiMdaaeaacaWG4bGaey4kaSIaaG4maaaacqGHRaWkdaGcaa % qaaiaaisdacaWG4baaleqaaaaa!4270! f\left( x \right) = \frac{{x + 9}}{{x + 3}} + \sqrt {4x} \) tại điểm x = 1 bằng:
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacqGH % sislcaaI2aaabaWaaeWaaeaacaWG4bGaey4kaSIaaG4maaGaayjkai % aawMcaamaaCaaaleqabaGaaGOmaaaaaaGccqGHRaWkdaWcaaqaaiaa % ikdaaeaadaGcaaqaaiaaisdacaWG4baaleqaaaaaaaa!44D0! f'\left( x \right) = \frac{{ - 6}}{{{{\left( {x + 3} \right)}^2}}} + \frac{2}{{\sqrt {4x} }}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaaIXaaacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacqGH % sislcaaI2aaabaWaaeWaaeaacaaIXaGaey4kaSIaaG4maaGaayjkai % aawMcaamaaCaaaleqabaGaaGOmaaaaaaGccqGHRaWkdaWcaaqaaiaa % ikdaaeaadaGcaaqaaiaaisdacaGGUaGaaGymaaWcbeaaaaGccqGH9a % qpdaWcaaqaaiaaiwdaaeaacaaI4aaaaaaa!475D! f'\left( 1 \right) = \frac{{ - 6}}{{{{\left( {1 + 3} \right)}^2}}} + \frac{2}{{\sqrt {4.1} }} = \frac{5}{8}\)