Answer
See below.
Work Step by Step
1. For $n=1$, we have $LHS=\frac{d(x)}{dx}=1$ and $RHS=(1)x^0=1=LHS$, thus P(1) is true.
2. Assume P(k) ($k\gt1$) is true, we have:
$\frac{d(x^k)}{dx}=kx^{k-1}$
3. For $n=k+1$, we have:
$LHS=\frac{d(x^{k+1})}{dx}=\frac{d(x^{k}x)}{dx}=x\frac{d(x^k)}{dx}+x^k
=xkx^{k-1}+x^k=(k+1)x^k=RHS$
4. Thus $P(k+1)$ is also true and we have proved the statement by mathematical induction.