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

I have two questions that i'm curious about.

1. If $f$ is differentiable real function on its domain, then $f'$ is Riemann integrable.

2. If $g$ is a real function with intermediate value property, then $g$ is Riemann integrable.

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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 $[-1,1]$.
Note that one can verify that $f'$, with $f$ is as above, has the Intermediate Value Property directly, rather than appealing to Darboux. –  David Mitra Dec 12 '12 at 14:30
1. A Riemann integrable function $f$ on an interval $[a,b]$ must be bounded on that interval. So if you take $f(x) = x^{\frac{3}{2}} \sin(\frac{1}{x})$ 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 $f(x) = \frac{1}{x} \sin(\frac{1}{x})$ on $[0,1]$ with $f(0) = 0$. 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.