Differentiability at a point
For
= \begin{cases}x^2,&\quad \textrm{x is rational}\\x^3,&\quad \textrm{x irrational}\end{cases})
I want to know where 
is differentiable/not differentiable. I know it is differentiable at 
.. but I kind of cheated and used the Squeeze Theorem to prove that.
I now want to that to show that is not differentiable at the only other point where it is continuous, i.e. at 
. I don't know how to show it explicitly in limit form, however.
Any help or tips are appreciated - thanks in advance!
Re: Differentiability at a point
The limit form requires that for any 
there is a 
, such that -{f(x+h)-f(x) \over h}|<\epsilon)
if 
.
However, if 
, there is always an h, for which this is not true (for any 
and any choice of f'(x)).
This is the case, since between any 2 different rational numbers, there is always an irrational number.
And between any 2 different irrational numbers, there is always a rational number.
Re: Differentiability at a point
Worded very well - thanks so much. I didn't think about using that form of limit.
