Answer
$f'(0)=0$ if $n \geq 2$
Work Step by Step
Apply the limit definition.
$f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$
and $f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)^n-x^n}{h}$
when $x=0$, then
$f'(0)=\lim\limits_{h \to 0}\dfrac{h^n-0}{h}=\lim\limits_{h \to 0}h^{n-1}$
when $n \geq 2$ , then we have $n-1 \geq 1$
Thus, $f'(0)=0$ if $n \geq 2$
