Mathematics Engineering: How do you prove the power rule?

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: $$\frac{d}{dx} x^r = \frac{d}{dx}\exp(r\ln x) = \exp(r\ln x)(r(\ln x)') = \frac{r}{x}\exp(r\ln x) = \frac{r}{x}(x^r) = rx^{r-1}.$$

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 $\frac{d}{dx}\ln x =\frac{1}{x}$. ofcourse this works if $x>0$.