Đạo hàm của hàm số \(y=\dfrac{x+1}{3^x}\) là
![]() | \(\dfrac{1}{3^x\ln3}\) |
![]() | \(\dfrac{1-(x+1)\ln3}{3^x}\) |
![]() | \(1-(x+1)\ln3\) |
![]() | \(\dfrac{\ln3-x-1}{3^x\ln3}\) |
Chọn phương án B.
\(\begin{aligned}
y'&=\dfrac{(x+1)'3^x-(x+1)\left(3^x\right)'}{3^{2x}}\\
&=\dfrac{3^x-(x+1)3^x\ln3}{3^{2x}}\\
&=\dfrac{3^x\left[1-(x+1)\ln3\right]}{3^{2x}}\\
&=\dfrac{1-(x+1)\ln3}{3^{2x}}.
\end{aligned}\)