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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?2012-12-29
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    @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.2012-12-29
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    As a part of learning the machinery of this website, I suggest the following exercises: (1) click the word *edit* under your post and insert the missing letter in the title. I don't like how **fuction** looks and sounds. (2) Although your post is quite readable as is, the formulas will look better if you enclose them in dollar signs $\$$. For example, you get $f'$ instead of f'. This will be important when you need to use more complex formulas. [Here's a TeX tutorial](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference)2012-12-29
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    @PavelM Oh, thanks. I'm going to learn it.2012-12-29

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