Again, let f be defined in a piecewise way by f(x)=x^2sin(1/x) when x!=0 and f(0)=0.
Certainly, when x!=0 the function is differentiable and continuous and the Product Rule and Chain Rule can be used to say that:
f'(x)=2x*sin(1/x)+x^2*cos(1/x)*(-1)x^{-2}
=2xsin(1/x)-cos(1/x) for x!=0.
But f is continuous everywhere (lim_{x->0}f(x)=0=f(0) by the Squeeze Theorem) and f'(0) happens to exist as well! Do a limit calculation:
f'(0)=lim_{h->0}(f(0+h)-f(0))/h=lim_{h->0}(h^2*sin(1/h)-0)/h
=lim_{h->0}hsin(1/h)=0, where the last equality follows by the Squeeze Theorem.
In other words, f'(0)=0.
Here's a graph of the function f (blue) and its derivative f' (red). Even though the derivative exists everywhere, it is not well-behaved near the origin. Not only does it have infinitely many oscillations as x->0, but the oscillations never decrease below 1 in amplitude (and lim_{x->0}f'(x) fails to exist so that f' is not continuous at x=0). On the other hand, the graph of the derivative contains the point (0,0) as well.
Pretty amazing, isn't it?!?!
Here a closer view near the origin and also making sure we can still see the blue curve. It's oscillating infinitely often as x->0 as well, though the oscillations are rapidly decreasing in amplitude as x->0.
