Inverse function theorem made easy

When you first learn calculus, you learn how to take derivatives of inverse functions essentially by "treating the derivative like a fraction,"

\begin{align} \label{eq:IFT} \dv{x}{y} = \frac{1}{\dv{y}{x}} \end{align}

where abuse of notation \(x(y)\) for the inverse of the function \(y(x)\) slightly obscures what's going on. Therefore, I present two very simple derivations of the relation Eq. \eqref{eq:IFT}.

In the first, we simply use the definition of an inverse function: \(y^{-1}(y(x)) = x\). Differentiating both sides gives

\begin{align} \label{eq:2} \eval{\dv{y^{-1}}{z}}_{z=y(x)} \dv{y}{x} = 1 \end{align}

which immediately gives the correct interpretation of the expression in Eq. \eqref{eq:IFT}, namely \(\dv{x}{y} = \eval{\dv{y^{-1}}{z}}_{z=y(x)}\). We see here that the deficiency is due to the notation. When we write \(\dv{x}{y}\), the \(y\) is a dummy variable whose value is not specified. The convention is that it is set to whatever value of \(y\) is implied by the rest of the expression. But here \(y\) is in fact a function so its value is the number \(y(x)\). The "correct" notation therefore is \(\dv{x}{y} = \dv{x}{y(x)}\), which we can't really use because \(x\) has two different meanings in the same expression: in the "numerator" it is a function, whereas in the "denominator" it is a number input to the function \(y\).

In the second derivation, we use the definition of the derivative as a local linear approximation to a function. To do so, we Taylor expand both \(y(x)\) and \(x(y)\)

\begin{align} \label{eq:localy} y(x) &= y(x_{0}) + \eval{\dv{y}{x}}_{x=x_0}(x-x_{0}) \\ \label{eq:localx} x(y) &= x(y_{0}) + \eval{\dv{x}{y}}_{y=y_{0}}(y - y_{0}). \end{align}

Equation \eqref{eq:localy} can be "solved for \(x\)" to yield

\begin{align} \label{eq:lagrangeinversion} x = x_{0} + \qty(\eval{\dv{y}{x}}_{x=x_{0}})^{-1}(y(x) - y(x_{0})) \end{align}

which we can compare with Eq. \eqref{eq:localx} to see that if \(x_{0} = x(y_{0})\) and \(y(x_{0}) = y_{0}\), then

\begin{align} \qty(\eval{\dv{y}{x}}_{x=x_{0}})^{-1} = \eval{\dv{x}{y}}_{y=y_{0}}. \end{align}

This equation is very close in spirit to the original Eq. \eqref{eq:IFT}, because here we have not committed on which variable \(y\) or \(x\) is a "function" of the other. Indeed the point geometrically is that they are both mutually functions of each other as long as we are looking closely enough in a local neighborhood of the point \((x_{0}, y_{0})\).