Đạo hàm của hàm số \(y=\log_{\frac{2}{5}}\left(-x^2+2x+1\right)\) là
\(y'=\dfrac{\ln5}{\left(1+2x-x^2\right)\ln2}\). \(y'=\dfrac{2\left(x+1\right)\ln5}{\left(1+2x-x^2\right)\ln2}\). \(y'=\dfrac{1}{2\left(1-x\right)\left(1+2x-x^2\right)\left(\ln2-\ln5\right)}\). \(y'=\dfrac{2\left(1-x\right)}{\left(1+2x-x^2\right)\left(\ln2-\ln5\right)}\). Hướng dẫn giải:Có \(y=\log_{\frac{2}{5}}\left(-x^2+2x+1\right)\) suy ra \(y'=\dfrac{\left(-x^2+2x+1\right)'}{\left(-x^2+2x+1\right).\ln\frac{2}{5}}\)\(=\dfrac{-2x+2}{\left(-x^2+2x+1\right)\left(\ln2-\ln5\right)}\).