Consider the function $g(x)=\frac{1}{x^n}$ where $n \in\Bbb N$, prove that g is differentiable.
I tried to use the definition,
Let $c \in\Bbb R$, then:
$\frac{g(x)-g(c)}{x-c}=\frac{\frac{1}{x^n}-\frac{1}{c^n}}{x-c}=\frac{c^n-x^n}{(x-c)(x^n \cdot c^n)}=\frac{(c-x)^n}{(x-c)(x^n \cdot c^n)}=-\frac{(c-x)^{n-1}}{(x^n \cdot c^n)}$
Hence $\lim_{x \to c} -\frac{(c-x)^{n-1}}{(x^n \cdot c^n)}=-\lim_{x \to c} \frac{(c-x)^{n-1}}{(x^n \cdot c^n)}=-\lim_{x \to c} \frac{(0)^{n-1}}{(c^n \cdot c^n)}=-\lim_{x \to c} \frac{0}{c^{2n}}=0$
But I dont think it's okay since I know the derivative of $\frac{1}{x^n} \neq 0$
So how should I do this then?
Thanks in advance,