What is an example that a function is differentiable but derivative is not Riemann integrable

Hints:

1) The function $f$ defined by $$f(x)=\cases{ x^2\sin(1/x^2),&$x\ne0$ \cr 0,&$x=0$}$$ is differentiable on $[-1,1]$; but its derivative is unbounded on $[-1,1]$.

2) Derivatives enjoy the Intermediate Value Property (by Darboux's Theorem).

1. A Riemann integrable function $f$ on an interval $[a,b]$ must be bounded on that interval. So if you take $$f(x)=\cases{x^{\frac{3}{2}} \sin(\frac{1}{x}),&$x\ne0$ \cr 0,&$x=0$}$$ on $[0,1]$, you can check that this is a continuous and differentiable function but with an unbounded derivative and so it is not integrable. You can even construct an example of a differentiable function whose derivative is bounded but is still not Riemann integrable - see Volterra's function.
2. Again, you take a function such as $$f(x)=\cases{\frac{1}{x} \sin(\frac{1}{x}),&$x\ne0$ \cr 0,&$x=0$}$$ on $[0,1]$. This is a discontinuous, unbounded function that satisfies the intermediate value property, but not Riemann integrable. A bounded example is given by the derivative of Volterra's function.