Use the chain rule to compute the derivative of an inverse function. Given a function $f(x)$, let $g=f^{-1}(x)$. Then define the composite function $f(g(x))$.
Then, by the Chain Rule, $(f(g(x)))'=f'(g(x)) g'(x)$.
However, since $f(x)$ and $g(x)$ are inverses, $f(g(x))=x$ and $(f(g(x)))'=(x)'=1$. So $1=f'(g(x)) g'(x)$.
If you solve for $g'(x)$, you get $$\frac{1}{f'(g(x))}$$ which is the derivative of an inverse function.
QED
Is the above proof correct and if so, can it be considered mathematically rigorous?
 A: Suppose that $f$ is differentiable on $\Bbb{R}$, and $g(x)=f^{-1}(x)$ for all $x\in\DeclareMathOperator{\ran}{ran}\ran(f)$. By definition of an inverse function, $f(g(x))=x$. If we take the derivative of both sides, we get
$$
f'(g(x)) \cdot g'(x)=1
$$
and rearranging this becomes
$$
g'(x)=\frac{1}{f'(g(x))}
$$
or
$$
(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))} \, .
$$
The problem with calling the above working a proof is that we haven't justified why
$$
(f \circ g)'(x)=f'(g(x)) \cdot g'(x) \, .
$$
You might be tempted to say "well, that's just the chain rule", but be careful. The chain rule is not simply the formula
$$
(f \circ g)'(x) = f'(g(x)) \cdot g(x) \, ,
$$
but rather the following statement:

If $g$ is differentiable at $x$, and $f$ is differentiable at $g(x)$,
then $f \circ g$ is differentiable at $x$, and $$ (f \circ
g)'(x)=f'(g(x)) \cdot g'(x) $$

So have the hypotheses of the chain rule been met? Since $f$ is a differentiable function, it is certainly true that $f$ is differentiable at $g(x)$. But we haven't shown that $g=f^{-1}$ is differentiable at $x$, and so we are unable to call the above working a proof. Just to give you an idea for how things might go wrong, consider what would happen if $f'(f^{-1}(x))=0$. Hence, all that we have shown above is that if $f^{-1}$ is differentiable at $x$, then
$$
(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))} \, .
$$
However, with more work we can prove the following theorem:

Let $f$ be a continuous one-to-one function defined on a certain
interval, and suppose that $f$ is differentiable at $f^{-1}(x)$, with
derivative $f'(f^{-1}(x))\neq 0$. Then, $f^{-1}$ is differentiable at
$x$, and $$ (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))} \, . $$

Because of this theorem, it turns out the only way in which things can go wrong is if $f'(f^{-1}(x))=0$. However, a lot more work is needed to show this than expected.
