Derivative of Inverse Function

Theorem
Let $$f$$ be a continuous real function whose inverse is $$g$$.

That is, let $$y = f \left({x}\right) \iff x = g \left({y}\right)$$.

If $$g^{\prime} \left({y}\right) \ne 0$$, then $$f^{\prime} \left({x}\right) = \frac 1 {g^{\prime} \left({y}\right)}$$.

Proof
Let $$f \left({x + \delta x}\right) = y + \delta y$$.

Then $$g \left({y + \delta y}\right) = x + \delta x$$ and hence:

$$ $$ $$

Style Note
Leibniz's notation for derivatives ($$\frac{dy}{dx}$$) allows for a particularly elegant statement of this rule: $$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$$ where $$\frac{dx}{dy}$$ is the derivative of $$x$$ with respect to $$y$$ and $$\frac{dy}{dx}$$ is the derivative of $$y$$ with respect to $$x$$.

However, do not interpret this to mean that derivatives can be treated as fractions, it simply is a convenient notation.