Let $f:{\mathbb R} \to {\mathbb R}$ be defined by $$f(x)= \begin{cases} x^2, & x \text{ is rational} \\ 0, & x \text{ is irrational} \end{cases} $$ Show that $f'$ is differentiable at $x=0$ and find $f'(0).$
Thus far I have,
If $x$ is rational: $$\lim_{x \to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \frac{x^2}{x} = \lim_{x \to 0} x = 0.$$
If $x$ is irrational, I have $\lim_{x \to 0} 0 = 0.$
How do I connect this to prove $f'$ is differentiable at $x=0,$ and show $f'(0)=0$?