## University Calculus: Early Transcendentals (3rd Edition)

$f'(0)=0$ if $n \geq 2$
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 $n \geq 2$