<s>[INST] Prove $1/x^n$ is differentiable 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,
 [/INST] You should evidently assume $x\neq 0$.
Then, for $|h|<\delta$  so that $x+h\neq 0$
$$\begin{align}
  \mathop {\lim }\limits_{h \to 0} \frac{{\dfrac{1}{{{{\left( {x + h} \right)}^n}}} - \dfrac{1}{{{x^n}}}}}{h} &= \mathop {\lim }\limits_{h \to 0} \frac{{\dfrac{{{x^n} - {{\left( {x + h} \right)}^n}}}{{{x^n}{{\left( {x + h} \right)}^n}}}}}{h} \cr 
   \\&= \mathop {\lim }\limits_{h \to 0} \frac{1}{{{x^n}{{\left( {x + h} \right)}^n}}}\frac{{{x^n} - {{\left( {x + h} \right)}^n}}}{h} \cr 
   \\&=  - \mathop {\lim }\limits_{h \to 0} \frac{1}{{{x^n}{{\left( {x + h} \right)}^n}}}\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^n} - {x^n}}}{h} \cr 
   \\&=  - \frac{1}{{{x^{2n}}}}\left( {\frac{d}{{dx}}{x^n}} \right) \cr 
   \\&=  - \frac{1}{{{x^{2n}}}}n{x^{n - 1}} \cr 
   \\&=  - n{x^{ - n - 1}} \end{align} $$
In fact, in a neighborhood where $g(x)\neq 0$, if $g(x)$ is differentiable, then so is $g(x)^{-1}$ and $$\frac d {dx}g(x)^{-1}=-\frac{g'(x)}{g(x)^2}$$
</s>