On the derivative of the inverse function of 1-variable I am looking for some example s.t. 


*

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

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

*$f^{-1}(y)$ 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)?
 A: 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).
The example I know is a bit messy to write down. Think about it, and then I can give more details if needed. Perhaps you could also be a bit more precise about what exactly the "locally" part of condition (2) means; I'm not sure that my example fits that requirement.
