# difficulty with understanding the continuity of the derivative

Define $$f(x)=x^3\cos\frac{1}{x}$$ for $$x\neq0$$ and $$f(0)=0$$.
show $$f^{'} (x)$$ is continuous at $$x=0$$ but not differentiable at $$x=0$$

I am studying Goldberg book and I am confused here.
the book defines : let $$f$$ be real valued function on an interval in real numbers. we say $$f$$ has a derivative at point $$c$$ in that interval if $$\lim_{x \to c} \frac{f(x)-f(c)}{x-c}$$ exists and we denote it by $$f^{'}(c)$$.
my first question is, does this definition means that in real analysis, we do not have to check the continuity of the function to decide wether a function is differentiable at all? is this enough to form this limit and if the limit exists we say f has derivative at that point?

my second question is : is there any technique to prove the continuity of the derivative? or do I need to used $$\epsilon-\delta$$ definition?
I am kinda confused how to wrap up this topic.

Here we get $$f'(x)=3x^{2}\cos (\frac 1 x)-x\sin (\frac 1 x)$$ for $$x \neq 0$$ and $$f'(0)=0$$. Now you have to check that $$f'(x) \to 0$$ as $$x \to 0$$ which is quite easy in this case since $$\sin$$ and $$\cos$$ are bounded functions.
I leave it to you to show using definition of derivative that that $$f''(0)$$ does not exist.