# Does differentiable function of bounded variation have bounded derivative?

I learned that $f$ is a function of bounded variation, when function $f$ is differentiable on $[a,b]$ and has bounded derivative $f'$.

What I want to know is converse part. If $f$ is differentiable on $[a,b]$ and $f$ is a function of bounded variation, Is derivative of $f$ bounded? I guess it's false, but i cannot find a counterexample. If it's true, please show me proof.

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Welcome to Math.SE! You are more likely to get effective answers here if you put in a bit more information about your effort to solve the problem. For example, do you know an example of a differentiable function with unbounded derivative on an interval? If yes, did you check if that function has bounded variation? – user53153 Dec 29 '12 at 17:54
@PavelM thanks for your attetion. I found some examples. for instance, f:=\sqrt{x} on [0,1] is a function of bounded variation because it's monotonic, but f has unbounded derivative. But actually, f is differentiable only on (a,b), not [a,b]. I'm finding counterexample whose domain of derivative is also closed interval, but it isn't going well. – gy6565 Dec 29 '12 at 18:32

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