On the derivative of the inverse function of 1-variable

I am looking for some example s.t.

1. $f:X\subset R \to R$ is a differentiable function at $x_0 \in X$,

2. $f'(x_0)\ne 0$ and inverse function $f^{-1}(y)$ exist locally at neighborhood of $f(x_0)$, and

3. $f^{-1}(y)$ is not differentiable at $f(x_0)$.

I think, one possible candidate is $f(x)=x^2 sin(1/x) - 100x$.

Then it satisfies the condition (1) and (2).

But I can't show condition (3).

Does it violate (3)?

Or if it is not the case, would you suggest me some example satisfying the conditions (1), (2), and (3)?

• Did you look at en.wikipedia.org/wiki/Inverse_function_theorem ? – Dirk Oct 10 '14 at 7:11
• Yes I saw it. I know that continuously differentiable function satisfy the inverse theorem. So, I think the counter example in my post should be differentiable but not continuously differentiable. – user29422 Oct 10 '14 at 8:35

No, your candidate will not work. If $f$ is continuous in an open interval containing $x_0$, then in order to be invertible it must be strictly increasing or strictly decreasing (which your candidate is), and then $f^{-1}$ will automatically be continuous and monotonous too. And in that situation, it's a one-line computation to prove directly from the definition that $(f^{-1})'(y_0)=\frac{1}{f'(x_0)}$ (where $y_0=f(x_0)$, of course). (The only thing that's really needed is that $f^{-1}$ is defined in a neighbourhood of $y_0$ and continuous at $y_0$.)
So if you want to violate (3), you will have to give up continuity. Of course, $f$ must be continuous at the point $x_0$ in order to be differentiable there, but it doesn't have to be continuous at all other nearby points. And in fact it's possible to construct a (globally) invertible $f$ with $f'(x_0)\neq 0$ such that $f^{-1}$ isn't even continuous at $y_0$ (and hence not differentiable there either).