Jtdylktuy

Let $pinmathbb R$, and let $f:mathbb Rtomathbb R$ be defined by
$$begin{cases}x^p&xinmathbb Q\0&xinmathbb Rsetminusmathbb Qend{cases}$$
For what values of $p$ is $f$ differentiable at $0$?

Would this just be a direct proof of differentiability (i.e. using the limit definition of the derivative)? Don't really know how to go about proving this rigorously.

Yes, with the definition of the derivative:

Let $q(x):= frac{f(x)-f(0)}{x-0}$ for $x ne 0$.

Then $q(x)=x^{p-1}$ if $x in mathbb Q$ and $q(x)=0$ if $x in mathbb R setminus mathbb Q$.

Case 1: $p=1$. Then $q(x)=1$ if $x in mathbb Q$. Can you see that $ lim_{x to 0}q(x)$ does not exist ?

Case 2: $p>1$. Then we have $|q(x)| le |x|^{p-1}$ for all $x$. Can you see that $ lim_{x to 0}q(x)=0$ ?

Case 3: $p<1$. It is now your turn to investigate the existence of $ lim_{x to 0}q(x)$ .

First you need continuity in $0$. If $p<0$ then the function is not continuous in zero since $$a_n={pi over n}\b_n={1over n}$$while both $a_n ,b_n$ tend to $0$ we have $$f(a_n)=0to 0\f(b_n)=({1over n})^{p}=n^{-p}to infty$$since $p<0$. Also $p=0$ is invalid since $f(a_n)=0$ and $f(b_n)=1$. Therefore we need $p>0$. After attaining this primary result, now we apply the definition of differentiability in $0$ for which$$f'(0)=lim_{xto 0} {f(x)over x}$$defining the same $a_n$ and $b_n$ we have $$f'(a_n)=0$$and $$f'(b_n)=lim_{xto 0}x^{p-1}=0$$which imposes that $p>1$ with the same argument. Therefore the final answer would be $p>1$.