# 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?

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$c^n-x^n\neq (c-x)^n$ if $n>1$. However $c-x$ is a factor of $c^n-x^n$. – Shane O Rourke Nov 13 '12 at 23:32
Stupid answer nvm. – Bob Nov 13 '12 at 23:34

Do this. $${g(x)- g(c)\over x - c} = {1/x^n - 1/c^n\over x - c} = {c^n-x^n\over x^nc^n(x-c)}.$$

Next, observe that by the Geometric Series Theorem $$x^n - c^n = (x - c)\sum_{k=0}^{n-1} x^k c^{n-1-k}$$

This allows you to cancel the $x-c$. Let $x\to c$ and you will be in business.

I will now reel this fish in. You now can combine our results to get

$${g(x)- g(c)\over x - c} = - {(x - c)\over x^n c^n}\sum_{k=0}^{n-1} x^k c^{n-1-k}/(x-c) = -{1\over x^n c^n}\sum_{k = 0}^{n-1}x^k c^{n-1-k} \to -{1\over c^{2n}}nc^{n-1} = {-n\over c^{n+1}}$$

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Okay, but then I find $-\frac{1}{c^{n+1}}$, so I miss a factor n in the numerator, but I dont know where I could have missed it then... – Bob Nov 13 '12 at 23:51
What happens to that as $x\to c$ (where $c\not = 0$? – ncmathsadist Nov 13 '12 at 23:54
To get the last equality on the top line I multiply top and bottom by $x^nc^n$. – ncmathsadist Nov 13 '12 at 23:55
suppose I already divided the (x-c) out, then I have left: $$-\frac{c^{n-1}}{c^n \cdot c^n}=-\frac{1}{c \cdot c^n}=\frac{1}{c^{n+1}}$$ I just dont understand from where you say that I'll be in business, cause I dont find the correct answer – Bob Nov 13 '12 at 23:59

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 $$\frac d {dx}g(x)^{-1}=-\frac{g'(x)}{g(x)^2}$$

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Peter, this puts two cool approaches on display here. – ncmathsadist Nov 13 '12 at 23:38