$D f^{-1}(x)= \frac{1}{f'(f^{-1}(x))}$

Question

Can someone explain what this rule means (with an example): $D f^{-1}(x)= \frac{1}{f'(f^{-1}(x))}$

I found this in the chapter of derivation but I can't understand the meaning

Try $f(x)=\tan{x}$ with inverse $f^{-1}(x)=\tan^{-1}(x).$ Both sides of the equation will be $1/(1+x^2)$ if done correctly. — user321120, Oct 24 '20 at 16:21
It's worth mentioning, you can rediscover it by starting at $f(f^{-1}(x))=x$ and then differentiating with the chain rule, $f'(f^{-1}(x)) * (f^{-1})'(x) = 1$. — Merosity, Oct 24 '20 at 16:46

score 1 · Answer 1 · answered Oct 24 '20 at 16:24

1

This is the derivation formula of the inverse function. Check this for more details. https://oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/inverseDeriv.html

answered Oct 24 '20 at 16:24

AlexanderGrey

392

score 0 · Accepted Answer · answered Oct 24 '20 at 16:26

When two sets are in bijection, they are basically the same set but written in two different languages. A bijection $f : A \to B$ between two sets $A$ and $B$ is a sort of dictionnary which translates $A$ into $B$. An object $a \in A$ is translated into $f(a) \in B$. The inverse bijection $f^{-1} : B \to A$ is just the reverse translation (and thus this is not extraordinary to have $f^{-1}(f(a)) = a$).

If $A$ and $B$ are intervals of the real line $\mathbb{R}$, one can ask, in the case of differentiable functions, how to have a translation of the differentiation properties. If $y = f(x)$, the chain rule shows the formula $(f^{-1})'(y) = \dfrac{1}{f'(x)}$. It says that the infinitesimal growth of the function $f^{-1}$ at a point $y = f(x)$ is the inverse of the infinitesimal growth of the function $f$ at $x$. You can visualize it by drawing the representative graphs of $f$ and $f^{-1}$ respectively: they are mirror-symmetric with respect to the line $y=x$, and the rate of the mirror-symmetric of a line is the inverse of the rate of the previous line.

$D f^{-1}(x)= \frac{1}{f'(f^{-1}(x))}$

2 Answers2