Proving differentiability I just had a question on proving differentiability by showing that the difference quotient exists. I understand in the case of a function like $f(x)=x^2$, where you end up with $((x+h)^2 - x^2)/h = 2x + h = 2x$ as h goes to infinity, but in the case of a function such as $1/x^n$, how do you address the "n" portion since you cannot expand and divide out the $h$ in the denominator?
 A: METHOD 1:  BRUTE FORCE USE OF THE DEFINITION OF THE DERIVATIVE
We can actually approach this with brute force using the binomial expansion.  We can write
$$(x+h)^n=x^n\left(1+n\frac hx+\frac{n(n-1)}{2!}\frac{h^2}{x^2}+\cdots+\frac{h^n}{x^n}\right)$$
Therefore,
$$\begin{align}
\frac{1}{(x+h)^n}-\frac{1}{x^n}&=\frac{1}{x^n}\left(\frac{1-\left(1+n\frac hx+\frac{n(n-1)}{2!}\frac{h^2}{x^2}+\cdots+\frac{h^n}{x^n}\right)}{\left(1+n\frac hx+\frac{n(n-1)}{2!}\frac{h^2}{x^2}+\cdots+\frac{h^n}{x^n}\right)}\right)\\\\
&=\frac{1}{x^n}\left(\frac{-\left(n\frac hx+\frac{n(n-1)}{2!}\frac{h^2}{x^2}+\cdots+\frac{h^n}{x^n}\right)}{\left(1+n\frac hx+\frac{n(n-1)}{2!}\frac{h^2}{x^2}+\cdots+\frac{h^n}{x^n}\right)}\right)\\\\
&=\frac{-\left(n\frac h{x^{n+1}}+\frac{n(n-1)}{2!}\frac{h^2}{x^{n+2}}+\cdots+\frac{h^n}{x^{2n}}\right)}{\left(1+n\frac hx+\frac{n(n-1)}{2!}\frac{h^2}{x^2}+\cdots+\frac{h^n}{x^n}\right)}\tag 1
\end{align}$$
Dividing $(1)$ by $h$ and letting $h\to 0$, we obtain the result $-\frac{n}{x^{n+1}}$ as expected.

METHOD 2:  USE OF THEOREMS 
Theorem 1:  The product rule.  $(fg)'=fg'+f'g$.
Theorem 2:  The quotient rule.  $(f/g)'=-fg'/g^2+f'/g$.
Theorem 3:  The derivative of $x$ is $1$.
We use induction to show that $(x^n)'=nx^{n-1}$ using Theorem $2$ and Theorem $3$.  
First observe that Theorem $3$ forms a base case.  
Then, we postulate that for some $k$, $(x^k)'=kx^{k-1}$ and examine $(x^{k+1})'$.  
Note that $(x^{k+1})'=(x\times x^{k})'$.  By Theorem 2, the product rule, we have $(x\times x^{k})'=x(x^k)'+x^k(x)'=x(kx^{k-1})+x^k=(k+1)x^k$.  
Therefore, by induction we have proven that $(x^n)'=nx^{n-1}$.
Next, we analyze $(1/x^n)'$.  Using Theorem 1 with $f=1$ and $g=x^n$ yields, $(1/x^n)'=-nx^{n-1}/x^{2n}+0=-n/x^{n+1}$ and  we are done!
A: We can still manipulate the definition purely algebraically: Let $x \neq 0$; let $n \geq 1$; note that
$$
h^{-1} \left( \frac{1}{(x+h)^{n}} - \frac{1}{x^{n}} \right) = h^{-1}\left( \frac{x^{n} - (x+h)^{n}}{(x^{2}+hx)^{n}} \right) = \frac{-\sum_{k=0}^{n-1}x^{k}(x+h)^{n-1-k}}{(x^{2}+hx)^{n}} \to -nx^{-n-1} 
$$
as $h \to 0$.
A: Rewriting in the form $$x^{\alpha}=e^{\alpha\ln(x)}$$ and the derivative is $$e^{\alpha\ln(x)}\frac{\alpha}{x}=\alpha x^{\alpha-1}$$
