Answer
$$ y^{\prime}=(x+1)^{x}\left[\frac{x}{x+1}+\ln (x+1)\right]$$
Work Step by Step
Given $$ y=(x+1)^{x} $$
So, we have:
\begin{aligned}
&y=(x+1)^{x} \\
&\Rightarrow \ln y=\ln (x+1)^{x} \\
&\Rightarrow \ln y=x \ln (x+1) \\
&\text{ differentiate with respect to } \ x,\\
&\Rightarrow \frac{y^{\prime}}{y}=\ln (x+1)+x \cdot \frac{1}{(x+1)}
\\&
\Rightarrow y^{\prime}=y\left[\ln (x+1)+ \frac{x}{(x+1)} \right]
\\&
\Rightarrow y^{\prime}=(x+1)^{x}\left[\frac{x}{x+1}+\ln (x+1)\right]
\end{aligned}
