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