Alexander Bass
This note is one of many taken during 2025

Calculus of Inverses

Let f(x),g(x)f(x), g(x) be functions that are defined and differentiable over some interval. Further, let

f(g(x))=xg(f(x))=x \begin{aligned} f(g(x)) &= x\\ g(f(x)) &= x\\ \end{aligned}

Such that g,fg,f are inverses.

Differentiate by chain rule.

f′(g(x))g′(x)=1 f'(g(x))g'(x) = 1 g′(x)=1f′(g(x)) g'(x) = \frac{1}{f'(g(x))}

This already has some neat results

f(x)=sin⁡(x)g(x)=sin⁡−1(x)ddxsin⁡−1(x)=1cos⁡(sin⁡−1(x)) \begin{aligned} f(x) &= \sin(x)\\ g(x) &= \sin^{-1}(x)\\ \frac{d}{dx} \sin^{-1}(x) &= \frac{1}{\cos(\sin^{-1}(x))} \end{aligned} f(x)=cos⁡(x)g(x)=cos⁡−1(x)ddxcos⁡−1(x)=−1sin⁡(cos⁡−1(x)) \begin{aligned} f(x) &= \cos(x)\\ g(x) &= \cos^{-1}(x)\\ \frac{d}{dx} \cos^{-1}(x) &= -\frac{1}{\sin(\cos^{-1}(x))} \end{aligned} f(x)=tan⁡(x)g(x)=tan⁡−1(x)ddxtan⁡−1(x)=1sec⁡2(tan⁡−1(x)) \begin{aligned} f(x) &= \tan(x)\\ g(x) &= \tan^{-1}(x)\\ \frac{d}{dx} \tan^{-1}(x) &= \frac{1}{\sec^2(\tan^{-1}(x))} \end{aligned}

These aren’t the standard forms you’re familiar with, but they are equivalent.

By this we can create a very annoying looking integral (for fun of course).

∫1sec⁡2(tan⁡−1(x))tan⁡−1(x)2 dx \int \frac{1}{\sec^2(\tan^{-1}(x))\tan^{-1}(x)^2} \,dx
This note is one of many taken during 2025