Đạo hàm của hàm số \(y=\dfrac{x+1}{4^x}\) là
\(y'=\dfrac{1-2\left(x+1\right)\ln2}{2^{2x}}\).\(y'=\dfrac{1+2\left(x+1\right)\ln2}{2^{2x}}\).\(y'=\dfrac{1-2\left(x+1\right)\ln2}{2^{x^2}}\).\(y'=\dfrac{1+2\left(x+1\right)\ln2}{2^{x^2}}\).Hướng dẫn giải:\(y=\dfrac{x+1}{4^x}\) có
\(y'=\dfrac{4^x\left(x+1\right)'-\left(x+1\right)\left(4^x\right)'}{\left(4^x\right)^2}\)
\(=\dfrac{4^x-\left(x+1\right)4^x\ln4}{4^{2x}}\)
\(=\dfrac{4^x\left[1-\left(x+1\right)\ln4\right]}{4^{2x}}\)
\(=\dfrac{1-\left(x+1\right)\ln2^2}{4^x}\)
\(=\dfrac{1-2\left(x+1\right)\ln2}{2^{2x}}\).