Inverse differentiation

We have some nice rules when it comes to differentiation, but for inverse functions, the mathematics comes a lot less easy when we look for non-first principles approaches, which motivates our exploration of inverse differentiation.

This is based on a core theorem below.

Let $f$ be differentiable and have an inverse on an interval $I$. Then for every value of $x$ at which $f^{\prime }(f^{-1}(x))\ne 0$, we can conclude that: $\frac{d}{dx}(f^{-1}(x))=\frac{1}{f^{\prime }(f^{-1}(x))}$

We recall that $f(f^{-1}(x))=x$ for all $x$ in the domain of $f^{-1}$. By differentiating implicitly, we can get the theorem’s result.

Inverse trigonometric functions

Using these results, we can derive the derivatives for inverse trigonometric functions like $\arcsin$, $\arccos$, and $\arctan$ fairly trivially. Let’s use $\arcsin$ as an example. From the theorem above, we get: $\frac{d}{dx}\arcsin (x)=\frac{1}{\cos (\arcsin (x))}$ From the trigonometric identity, ${\cos }^2(x)+{\sin }^2(x)=1$, we rearrange to get: $\cos (x)=\sqrt{1-{\sin }^2(x)}$ Then for our input, we put in $\arcsin (x)$ and simplify: $\cos (\arcsin (x))=\sqrt{1-{\sin }^2(\arcsin (x))}$ $\cos (\arcsin (x))=\sqrt{1-x^2}$ So our final derivative, as we know, is: $\frac{d}{dx}\arcsin (x)=\frac{1}{\sqrt{1-x^2}}$

Simple applications

As an example, let $f(x)=x^3-3x^2+5x+1$, $g(x)=f^{-1}(x)$, and let’s try to find $g^{\prime }(4)$. By theorem: $\frac{d}{dx}(f^{-1}(x)){|}_{x=4}=\frac{1}{f^{\prime }(f^{-1}(4))}$ Since we need to find $f^{-1}(4)=k$, where $k\in \mathbb{R}$. Then the point $(4,k)$ is on the graph of $f^{-1}$, and by extension the point $(k,4)$ is on the graph of $f$, since the ranges and domains are swapped.

So we find $f(k)=4$, instead, and we observe that $k=1$ gives us a solution. From here, the rest of the exercise is fairly trivial.