Answer
$$y'=(x+1)^x\Big(\ln(x+1)+\frac{x}{x+1}\Big)$$
Work Step by Step
$$y=(x+1)^x$$
We note that $(x+1)^x=e^{\ln(x+1)^x}=e^{x\ln(x+1)}$
Therefore, $$y=e^{x\ln(x+1)}$$
The derivative of $y$ is: $$y'=(e^{x\ln(x+1)})'=e^{x\ln(x+1)}(x\ln(x+1))'$$
$$y'=e^{x\ln(x+1)}(\ln(x+1)+x\times\frac{1}{x+1}\times(x+1)')$$
$$y'=e^{x\ln(x+1)}\Big(\ln(x+1)+\frac{x}{x+1}\Big)$$
$$y'=(x+1)^x\Big(\ln(x+1)+\frac{x}{x+1}\Big)$$
