# Q: Proving differentiability at a point for a function based on differentiability of its inverse.

Preface. Here is the problem I am working on:

Suppose $$I$$,$$J$$ are intervals and a monotone onto $$f:I\to J$$ has an inverse $$g:J\to I$$. Suppose $$x\in I$$ and $$y:=f(x)\in J$$, and that $$g$$ is differentiable at $$y$$. Prove:

1. If $$g'(y)\ne 0$$, then $$f$$ is differentiable at $$x$$.
2. If $$g'(y) = 0$$, then $$f$$ is not differentiable at $$x$$.

I've been working on this assignment for so long now that I'm at my breaking point. I feel like all I need is a little push in the right direction, that is what kinds of things should I be thinking about to get started with this? I don't want a full solution, just some suggestions and pointers.

Hint:

By hypothesis,

$$\lim_{h\to0}\frac{g(x+h)-g(x)}h=g'(x)$$ meaning that

$$\forall \epsilon>0:\exists\delta:\forall|h|<\delta\implies\left|\frac{g(x+h)-g(x)}h-g'(x)\right|<\epsilon.$$

As $$g$$ is invertible, $$x=g^{-1}(y)$$ and $$x+h=g^{-1}(y+l)$$ for some $$l$$, and the last condition writes

$$\left|\frac{l}{g^{-1}(y+l)-g^{-1}(y)}-g'(x)\right|<\epsilon.$$

Now by choosing $$\delta'=\min(|g(x)-g(x-h)|,|g(x+h)-g(x)|)$$ (noting that $$\forall \epsilon:\exists \delta'$$) and by monotonicity, the inequation holds for all $$|l|<\delta'$$.

The proves that

$${\lim_{l\to0}}\dfrac l{g^{-1}(y+l)-g^{-1}(y)}=\frac1{{\lim_{l\to0}}\dfrac{g^{-1}(y+l)-g^{-1}(y)}l}=\frac1{{g^{-1}}'(y)}=g'(x),$$ unless $$g'(x)=0$$.

• Thank you for the hint. I'll try to use this as I think about this problem more. – MrStormy83 Sep 16 at 20:47

For the sake of argument lets say that $$g$$ is differentiable at $$y_0 = f(x_0)$$ $$g^\prime(y_0) = \lim_{y\to y_0}\frac{g(y)-g(y_0)}{y-y_0}$$ Now we are going to do a substitution $$x=g(y)$$. If we remember that the inverse of a monotone function is continuous we can see that $$\lim_{y \to y_0}g(y) = g(y_0)$$. So according to the substitution theorem

$$g^\prime(y_0) = \lim_{x\to x_0}\frac{x-x_0}{f(x)-f(x_0)}$$ Substitution theorem implies the existence of the limit on the RHS. If $$g^\prime(y_0)\neq0$$ then

$$\frac{1}{g^\prime(y_0)} = \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0} = f^\prime(x_0)$$ In case $$g^\prime(y_0) = 0$$ the limit in the second equality is equal to $$0$$, so the limit in the third equation cant converge.

• This is extremely helpful. I appreciate the pointer, I think I can come up with a reasonable proof based on this information. – MrStormy83 Sep 16 at 20:46
• Glad I could be of help :) – Boxonix Sep 17 at 4:44