Derivatives of inverses

Question

So I know that if $f(3)=4$, $f^{-1}(4)=3$ for all $x,y$ where the functions are defined.. but does the same rule apply to derivatives? i.e. if $f'(3)=4$, does $(f^{-1})'(4)=3$ ?

Answer 1

No, you can come up with a counterexample yourself, as this should almost never hold.

Take $f:(0,\infty)\to (0,\infty$, $f(x)=x^2$. Then $f^{-1}: (0,\infty)\to (0,\infty), x\mapsto \sqrt{x}$.

Now calculate, for example, $f'(2)=4$. But $(f^{-1})'(4)=\frac{1}{2\sqrt{4}}=\frac14$.

Answer 2

No. Take $$f:\mathbb{R}_+\to\mathbb{R}_+~~~~~~~f:x\mapsto x^2$$ Its inverse is $$f^{-1}(x)=\sqrt{x}$$ Observe that $$f'(x)=2x$$ So $$f'(3)=6$$ But $$(f^{-1})'(x)=\frac{1}{2\sqrt x}$$ Thus $$(f^{-1})'(6)=\frac{1}{2\sqrt{6}}\neq 3$$ In response to some comments, notice that if we reverse the order, then $$(f')^{-1}(6)=3$$ Since $$(f')^{-1}:2x\mapsto x\implies (f')^{-1}(x)=\frac{x}{2}$$ But this is rather obvious, taking e.g $f^{-1}=g$ all this is saying is $$x=(g^{-1}\circ g)(x)$$ Which is the definition of an inverse. What you can say in general is as follows: Let $\phi$ be differentiable and invertible. Let $\mathrm{D}$ be the differential operator. Then by definition of inverse $$\phi^{-1}(\phi(x))=x$$ Taking a derivative, $$\mathrm{D}\left(s\mapsto \phi^{-1}(\phi(s))\right)(x)=\mathrm{D}(s\mapsto s)(x)$$ Using our chain rule, $$\mathrm{D}(\phi^{-1})(\phi(x))\phi'(x)=1$$ So $$\mathrm{D}(\phi^{-1})(\phi(x))=\frac{1}{\phi'(x)}$$ Or, letting $\phi(x)=y\implies x=\phi^{-1}(y)$ and switching back to the prime notation we have $$(\phi^{-1})'(y)=\frac{1}{\phi'(\phi^{-1}(y))}$$ This formula is nice, but actually rather useless. Basically it says in order to compute the derivative of the inverse at some input, say $x$, one first needs to compute $\phi^{-1}(x)$. But, if we have an easy way of computing $\phi^{-1}$, then in practice we also already have an easy way to compute $(\phi^{-1})'$.

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Key takeaway: Inverse functions are a real pain in the ass.

Answer 3

Not so. We can see by a rough graph of its inverse function, here particularly $ \sin , \sin^{-1}$

For a function and its inverse.. due to a reflection about black straight line $x=y$ coordinates are swapped but the slopes cannot be swapped in same way generally.

$$ \text{ If} f(u)=v, \text{then} (f^{−1})′(v)=1/f′(u). $$

Due to symmetry w.r.t. line $x=y$ if $ \phi $ is slope for f, then for $f^{-1} $ the slope goes to its complement at corresponding mapped points on the red/blue lines. This is simply because

$$ \frac{dy}{dx} \to \frac{1}{{dx}/{dy}}. $$

$$ \tan(..) \to 1/\tan(..) = \cot(..)= \tan( \pi/2- ...)$$

$$\phi \to \pi/2-\phi $$