# A function with a non-zero derivative, with an inverse function that has no derivative.

While studying calculus, I encountered the following statement: "Given a function $$f(x)$$ with $$f'(x_0)\neq 0$$, such that $$f$$ has an inverse in some neighborhood of $$x_0$$, and such that $$f$$ is continuous on said neighborhood, then $$f^{-1}$$ has a derivative at $$f(x_0)$$ given by: $${f^{-1}}'(x_0)=\frac{1}{f'(x_0)}$$

My questions is - why does $$f$$ have to be continuous on a whole neighborhood of $$x_0$$ and not just at $$x_0$$? Is there some known counter-example for that?

• Welcome to MSE. Nice first question! – José Carlos Santos Jan 6 '19 at 22:39
• Example 1 in this post shows that continuity is not necessary (for pointwise differentiability of the inverse). – user21820 Jan 7 '19 at 13:52

The suggestion in the title isn't how it'll work. Instead of having an inverse that doesn't have a derivative, we'll fail to have a continuous inverse. Also, the required condition for the theorem isn't just that $$f$$ is continuous on an interval - it's that $$f'$$ is continuous on an interval around the key point.

Example: $$f(x)=\begin{cases}x+2x^2\sin\frac 1x&x\neq 0\\0&x = 0\end{cases}$$. This $$f$$ is differentiable everywhere, with derivative $$1$$ at zero, but it doesn't have an inverse in any neighborhood of zero. Why? Because it isn't monotone on any neighborhood of zero. We have $$f'(x)=1+4x\sin\frac1x-2\cos\frac1x$$ for $$x\neq 0$$, which is negative whenever $$\frac1x\equiv 0\mod 2\pi$$. We can find a one-sided inverse $$g$$ with $$f(g(x))=x$$, but this $$g$$ will necessarily have infinitely many jump discontinuities near zero.

The calculation of the derivative of $$f^{-1}$$ is just an application of the chain rule. The real meat of the inverse function theorem is the existence of a differentiable inverse.

• Yes I can see it now. Thank you very much! – Ran Kiri Jan 6 '19 at 23:01

First off, any function has an inverse "at $$x_0$$" because we just assign $$f(x_0)$$ the value $$x_0$$; it's really meaningless to talk at an inverse existing at a point. We need a whole neighbourhood because then we can use the derivative, which is defined by a limit, so we must be able to "approach" $$x_0$$ arbitarily closely.

I think that as long as $$f^{-1}$$ is well-defined on a neighborhood of $$f(x_0)$$, and continuous at $$f(x_0)$$, there is no issue.

Indeed, $$f(f^{-1}(f(x_0)+h))=f(x_0)+h$$ so $$h=f(f^{-1}(f(x_0)+h))-f(x_0) \sim f’(x_0)(f^{-1}(f(x_0)+h)-x_0)$$, and the conclusion (of differentiability and value of the derivative) follows.

The continuity condition is not necessary. It's enough that $$f$$ be injective on some neighborhood. This said, if your function has a sequence of jump discontinuities near $$x_0$$, you might have that there is no open interval $$U$$ around $$x_0$$ for which $$f(U)$$ is also an interval. This means that $$f^{-1}$$ might be defined on a strange domain, though we can still technically differentiate it to get the desired result.

Formally, the statement you would need to prove is the following:

Let $$A$$ and $$B$$ be subsets of $$\mathbb R$$ and $$f:A\rightarrow B$$ and $$g:B\rightarrow \mathbb R$$. Suppose that $$x_0\in A$$ is an accumulation point of $$A$$ and $$f(x_0)$$ is an accumulation point of $$B$$. Then,

• If two of the derivatives $$f'(x_0)$$ and $$g'(f(x_0))$$ and $$(f\circ g)'(x_0)$$ exist and are non-zero, the third exists as well.

• If all of the derivatives exist, then $$(f\circ g)'(x_0)=f'(x_0)\cdot g'(f(x_0)).$$

One you have this statement, you can apply it to a pair where we take $$g=f^{-1}$$. Note that we can make this work even if $$f$$ isn't defined on an interval around $$x_0$$ - it's okay as long as we have enough points to define the relevant limit towards $$x_0$$.

Granted, it is a bit unusual to talk about derivatives on sets that aren't open, but there's no technical limitations preventing it, though the proof of the suggested lemma is a pain.