Help with proof involving derivatives

Question

I need to show thath for $f(x)=\frac{1}{x^{n}}$ $n\in\mathbb{Z^{+}}$, $f'(x)=\frac{-n}{x^{n+1}}$. So this is what I've got so far $f'(x)=\lim_{h\to0} \frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\frac{1}{(x+h)^n}-\frac{1}{x^n}}{h}=\lim_{h\to0}\frac{\frac{(x^n)}{(x+h)^n(x^n)}-\frac{(x+h)^n}{(x+h)^n(x^n)}}{h}$.So what I'm having trouble with is how to cancel that $h$. Since $n\in\mathbb{Z}^{+}$ The term is the binomial expansion $(x+h)^n=$ $h=0$ So how do I cancel that $h$ before calculating the limit?. Thanks a lot in advance!.

Answer 1

You'll just need to tough it out and expand the first few terms of the binomial. The required cancellations will happen. That is, write, for some $n$,

$$ (x+h)^n = \sum_{j = 0} ^n \binom{n}{j} x^j h^{n-j} $$

Try some test cases then get the general form.

Answer 2

The proof is much simpler by induction. Here's a sketch of the inductive step: $$\biggl(\frac1{x^{n+1}}\biggr)'=\biggl(\frac1{x^n}\cdot\frac1x\biggr)'=\frac{-n}{x^{n+1}}\cdot\frac1x+\frac1{x^n}\cdot\frac{-1}{x^2}=-\frac{n}{x^{n+2}}-\frac{1}{x^{n+2}}=\frac{n+1}{x^{n+2}}.$$