Mathematics Engineering: How do you prove the power rule? Consider the real-valued function  $f(x)=x^r$ where $r$ is a real number.
(1) For what values of $x$ and $r$, is this function differentiable?
(2) How do you prove the power rule, when $f(x)$ is differentiable?
It would be nice if the proof is accessible to a 12th grade student.
EDIT: Perhaps, I have not been very clear about my question.  I am not satisfied with the "textbook proofs" I have seen.  Please established the existence (of the derivative) first before proving the power rule.  That is, please consider first question first.  This is the reason why I call the question a mathematical engineering question.  (If we can we bring high level math down to Grade 12 level, without loosing precision, then I call it good mathematical engineering.)
 A: Let's prove the power rule for an arbitrary real number. So $r$ can be an integer, rational, irrational. We make use of the fact that $e^{r\ln x}=x^r$.
$$
\begin{align*}
\frac{\text{d}}{\text{dx}} (x^r) & = \frac{\text{d}}{\text{dx}} \left(e^{r\ln x}\right)\\
&= e^{r\ln x}  \frac{\text{d}}{\text{dx}}\left(r\ln x \right)\\
& = e^{r\ln x}  \left(\frac{r}{x}\right)\\
& =rx^{r-1}.
\end{align*}
$$
where we have used the fact that $\frac{\text{d}}{\text{d}x}\ln x =\frac{1}{x}$. Of course this works if $x>0$.
A: For positive integer $r$, we can define $x\mapsto x^r$ for all $r$, and the formula follows from the definition of derivative and the Binomial Theorem:
$$\begin{align*}
\frac{d}{dx} x^r &= \lim_{h\to 0}\frac{(x+h)^r - x^r}{h} = \lim_{h\to 0}\frac{x^r + rx^{r-1}h + h^2(\text{some factors}) - x^r}{h}\\
&= \lim_{h\to 0}\frac{rx^{r-1}h + h^2(\text{some factors})}{h} = \lim_{h\to 0}\left(rx^{r-1} + h(\text{some factors})\right)\\
&= rx^{r-1} + 0 = rx^{r-1}
\end{align*}$$
It is continuous for all $x$, since it is differentiable for all $x$.
For negative integers $r= -n$ with $n$ a positive integer, continuity at all $x\neq 0$ follows because $x^n$ is continuous at all $x$. For differentiability, we know it is differentiable at all places where it is defined, because it is the quotient of two differentiable functions (the constant function $1$, and the function $x\mapsto x^n$, which we just proved is differentiable). To find the formula, we use the Product Rule:
$$\begin{align*}
0 &= \frac{d}{dx} 1 = \frac{d}{dx}x^{-n}x^n = (x^{-n})'x^n + (x^{-n})(x^n)'\\
&= (x^{-n})'x^n + x^{-n}(nx^{n-1}) = (x^{-n})' + nx^{-1}.
\end{align*}$$
Solving for $(x^{-n})'$ we obtain 
$$\frac{d}{dx} x^{-n} = -\frac{nx^{-1}}{x^n} = -nx^{-n-1} = rx^{r-1},$$
yielding the power rule. (We could also use the Quotient Rule to get differentiability of $\frac{1}{x^n}$). 
For rational $r=\frac{p}{q}$ with $p$ and $q$ in reduced terms, the definition is that $x^r = (x^p)^{1/q}$. For $q$ odd, this is defined for all $x$ and is continuous (it's the composition of $x^p$ and the inverse of $x^q$, which is continuous on all $x$), but differentiable only at $x\neq 0$; differentiability of $x\mapsto x^{1/q}$ follows by the Inverse Function Theorem, and differentiability of $x^{p/q}$ now follows because it is a composition of differentiable functions, so the proof of the Chain Rule shows that it is differentiable; for $q$ even, this is defined for all $x\geq 0$ and is continuous there, but only differentiable at $x\gt 0$ (the tangent at $x=0$ is vertical). Differentiability elsewhere again follows by the Inverse Function Theorem and the Chain Rule (which in fact proves that the composition is differentiable).
For $x^{1/q}$ with $q$ and integer, we use the Chain Rule:
$$1 = \frac{d}{dx}x = \frac{d}{dx}(x^{1/q})^q = q(x^{1/q})^{q-1}(x^{1/q})',$$
so solving for $(x^{1/q})'$ we get
$$\frac{d}{dx}x^{1/q} = \frac{1}{q}x^{(1-q)/q} = \frac{1}{q}x^{(1/q) - 1}.$$
Then for $x^{p/q}$ we have, again by the Chain Rule, 
$$\frac{d}{dx}x^{p/q} = \frac{d}{dx}(x^p)^{1/q} = \frac{1}{q}(x^p)^{(1/q)-1}(x^p)' = \frac{p}{q}(x^{p})^{(1/q)-1}x^{p-1} = \frac{p}{q}x^{(p/q)-p+p-1} = \frac{p}{q}x^{(p/q)-1}.$$
For irrational exponent $r$ we only define $x^r$ for $x\gt 0$, and we define it $x^r = \exp(r\ln x)$. For $x\gt 0$ the function $x\mapsto \ln x$ is differentiable; hence $x\mapsto r\ln x$ is differentiable. The exponential is differentiable everywhere, so the commposition $x\mapsto \exp(r\ln x)$ is differentiable everywhere that it is defined, again by the Chain Rule. To get the formula, we can just use the Chain Rule again:
$$\frac{d}{dx} x^r = \frac{d}{dx}\exp(r\ln x) = \exp(r\ln x)(r(\ln x)') = \frac{r}{x}\exp(r\ln x) = \frac{r}{x}(x^r) = rx^{r-1}.$$
A: That it is true of nonnegative integer values of $r$ can be proved by any of several methods.


*

*One of those uses mathematical induction and the power rule:
$$
\frac{d}{dx} x^{r+1} = \frac{d}{dx} (x^r\cdot x) = x\frac{d}{dx} x^r + x^r \frac{d}{dx} x = x(rx^{r-1}) + x^r\cdot 1 = (r+1)x^r.
$$

*Another uses the definition of derivative directly:
$$
\frac{d}{dx} x^r = \lim_{w\to x} \frac{w^r-x^r}{w-x} = \lim_{w\to x}\frac{(w-x)(w^{r-1}+w^{r-2}x + w^{r-3}x^2 + \cdots + x^{r-1})}{w-x}
$$
$$
= \lim_{w\to x} (w^{r-1}+w^{r-2}x + w^{r-3}x^2 + \cdots + x^{r-1})
$$
$$
= \underbrace{x^{r-1}+\cdots+x^{r-1}}_{r\text{ terms}}.
$$

*Another uses the binomial theorem and the definition of the derivative.  There's a bit of overkill in that one, since the binomial expansion gives you all $r+1$ coefficients where you need only the first two.
For positive $x$ and real $r$ (non-integer and not necessarily positive), you can use logarithmic differentation, provided you know the chain rule and how to differentiate the natural exponential and logarithmic functions:
$$
\frac{d}{dx} x^r = \frac{d}{dx} e^{r\log_e x} = e^{r\log_e x} \frac{r}{x} = x^r\cdot \frac{r}{x}.
$$
For negative integer values of $r$, you can use the reciprocal rule:
$$
\frac{d}{dx}x^r = \frac{d}{dx} \frac{1}{x^{-r}} = \frac{-\frac{d}{dx}x^{-r}}{x^{-2r}} = \frac{rx^{-r-1}}{x^{-2r}} = rx^{r-1}.
$$
It's also true of arbitrary real values of $r$ even when $x$ is not positive.  I'll think about the quickest way to prove that.  Except that it is not true when $x=0$ and $r<0$, since that involves dividing by $0$.
A: Prove power rule from first principle via binomial theorem and taking leading order term, now for negative exponents, we can use a trick. Consider:
$$ x^k \cdot x^{-k} = 1$$
The above identity holds for all $ x \in R -{0}$, differentiate it:
$$ kx^{k-1} x^{-k} + x^k \frac{d}{dx} x^{-k} =0$$
$$  \frac{d}{dx} x^{-k} =  -\frac{k}{x^{k+1}}$$
So, without any generalized binomial theorem, a student who knows binomial theorem + product rule would be able to do this :^)
It is definitely differentiable for all $r \in \mathbb{R}$ and all x in R except for case when $r$ is negative ; then x can't be zero
A: Edited to include absolute values.
Write $y=x^r$. Then $\log|y|=r\log|x|$. Then implicit differentiation gives $\displaystyle\frac{y'}{y}=r\frac{1}{x}$, or $\displaystyle y'=r\frac{y}{x} = r\frac{x^r}{x} = rx^{r-1}$. This works for all real numbers $r$, and so $f(x)=x^r$ is differentiable for all $r$ and for all $x$ for which $x^r$ is defined, unless $r-1<0$ in which case $f$ is not differentiable at $0$.
Of course, this assumes you already know that $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x} \log(x) = \frac{1}{x}$.
