The function, for x not 0, is \(\displaystyle f(x)= x+ x^2 sin(1/x)\) (this does not exit for x= 0 but the limit, as x goes to 0, is 0 so the given function is continuous at x= 0). It's derivative is \(\displaystyle f'(x)= 1+ 2x sin(1/x)+ x^2 cos(1/x)(-1/x^2)= 1+ 2x sin(1/x)- cos(1/x)\). That does not exist for x= 0 nor does it have a limit as x goes to 0.
The derivative of f at x= 0 is given by \(\displaystyle \lim_{h\to 0}\frac{f(0+h)- f(0)}{h}= \lim_{h\to 0}\frac{h+ h^2 sin(1/h)- 0}{h}= \lim_{h\to 0} 1+ h sin(1/h)= 1\).
(c) The function, f, is differentiable at x= 0 but the derivative is not continuous there.
(One can show that, while the derivative of a function is not necessarily continuous on a given interval, it must satisfy the "intermediate value property": "if f'(a)= P and f'(b)= Q, then f' must take on every value between P and Q for some x between a and b".