Inverse functions
Here I show a basic way to find the derivative of an inverse function and using that, finding the inverse function itself. For now confine ourselves to functions of one real variable.
Let \(y=f(x)\). Then \(y'=\frac{dy}{dx}=f'(x)\) and \(f^{-1}(y)=x\), wherever the inverse is well-defined and injective. Then,
\[\begin{align} & x=f^{-1}(y)=f^{-1}(f(x)) \nonumber \\ \implies & 1=\frac{d (f^{-1}(f(x)))}{df(x)}\frac{df(x)}{dx} \nonumber \\ \implies & \frac{d (f^{-1}(y))}{dy}=\frac{1}{f'(x)} \nonumber \end{align}\]Examples
Without knowing how to denote or name an inverse function a priori, we will use the designation \(\mathrm{inv}(y)\).
Example 1. Let \(y=e^x\). Then,
\[\begin{align} & \frac{d (f^{-1}(y))}{dy}=\frac{1}{f'(x)}=\frac{1}{e^x}=\frac{1}{y} \nonumber \\ \therefore \quad & \mathrm{inv}(y)=\int\frac{1}{y}\,dy \nonumber \end{align}\]We call this integral \(\ln(y)+c\), with \(c\) being the integration constant. Now when \(y=1\), \(\mathrm{inv}(y)\), i.e., \(x\), is zero. And we know that \(y=1 \implies \ln(1)=0\). So, \(c=0\).
Example 2. Let \(y=\sin(x)\), \(x\in[-\frac{\pi}{2},\frac{\pi}{2}]\). Then \(f'(x) = \cos(x)\). We know \(\cos(x) = \sqrt{1 - \sin^2(x)}\). We take the positive root because cosine is non-negative on our domain. Substituting \(y = \sin(x)\) yields \(f'(x) = \sqrt{1 - y^2}\).
\[\begin{align} & \frac{d (f^{-1}(y))}{dy}=\frac{1}{f'(x)}=\frac{1}{\sqrt{1 - y^2}} \nonumber \\ \therefore \quad & \mathrm{inv}(y)=\int\frac{1}{\sqrt{1 - y^2}}\,dy \nonumber \end{align}\]This integral defines the arcsine function, \(\arcsin(y) + c\). Here, \(y=0 \implies x=0\), and \(\arcsin(0) = 0\), so \(c=0\).
Example 3. Let \(y=\tanh(x)\). The derivative is \(f'(x) = \mathrm{sech}^2(x)\). Using the identity \(\mathrm{sech}^2(x) = 1 - \tanh^2(x)\), we can substitute \(y = \tanh(x)\) to express the derivative in terms of \(y\), yielding \(f'(x) = 1 - y^2\).
\[\begin{align} & \frac{d (f^{-1}(y))}{dy}=\frac{1}{f'(x)}=\frac{1}{1 - y^2} \nonumber \\ \therefore \quad & \mathrm{inv}(y)=\int\frac{1}{1 - y^2}\,dy \nonumber \end{align}\]This integral defines the inverse hyperbolic tangent function, \(\mathrm{artanh}(y) + c\). Since \(y=0 \implies x=0\), and \(\mathrm{artanh}(0) = 0\), we find that \(c=0\).