Theorem: Derivative of an Inverse Function

Theorem. If $y=f(x)$ is a differentiable function of $x$ with $\dfrac{dy}{dx}\ne0$, and $x=f^{-1}(y)$ exists, then $x=f^{-1}(y)$ is a differentiable function of $y$, and $\dfrac{dx}{dy}=\dfrac{1}{dy/dx}\qquad\left(\text{equivalently } \dfrac{d}{dy}[f^{-1}(y)]=\dfrac{1}{f'(x)}\right).$

Proof 1 (from increments). Let $x$ change by a small $\delta x\ne0$, producing a corresponding $\delta y\ne0$ in $y$. From the algebraic identity $\dfrac{\delta x}{\delta y}\cdot\dfrac{\delta y}{\delta x}=1$ (valid whenever $\delta y\ne0$), we get $\dfrac{\delta x}{\delta y}=\dfrac{1}{\delta y/\delta x}$. As $\delta x\to0$ we also get $\delta y\to0$ (since $y=f(x)$ is continuous), so taking the limit gives $\displaystyle\lim_{\delta y\to0}\dfrac{\delta x}{\delta y}=\dfrac{1}{dy/dx}$, which exists and is finite because $\dfrac{dy}{dx}\ne0$. Hence $x$ is differentiable in $y$ with $\dfrac{dx}{dy}=\dfrac{1}{dy/dx}$. …