cho y= log x+1 x. tính y'(1) . A. 0 b. 1/ln2

* Hướng dẫn giải

Đáp án B

$$\begin{gathered} y={\log }_{x+1}x=\frac{\ln x}{\ln (x+1)} \\ y\text{'}=\frac{{\displaystyle \frac{1}{x}}\ln (x+1)-\ln x.{\displaystyle \frac{1}{x+1}}}{{\left[\ln \right(x+1\left)\right]}^2}=\frac{(x+1)\ln (x+1)-x\ln x}{x(x+1){\left[\ln \right(x+1\left)\right]}^2} \\ \Rightarrow y\text{'}\left(1\right)=\frac{2\ln 2}{2{\left(\ln 2\right)}^2}=\frac{1}{\ln 2} \end{gathered}$$