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If a function is differentiable and monotone on the interval $(a, b)$, then its derivative is also monotone on $(a, b)$.

How do you prove this statement is wrong?

Can you please provide an example?

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This thing


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+1 just because it's so funny. – Git Gud Jan 13 '13 at 20:23
It's 2-valued at one interval. Need eraser. – alancalvitti Jan 13 '13 at 20:24
We need more pictures on MSE :) – StuartHa Jan 13 '13 at 21:21

Consider $f(x)=x^3$ on $[-1,1]$. Then it is clearly monotone, but $f'(x)=3x^2$ which is clearly not monotone.

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Google search has improved! search for x^3, 3*x^2 – user13107 Jan 14 '13 at 1:16

No, it shouldn't.

Explanation of this fact can be based on the convexity of the function, which depends on the second derivative of the function.

In other words: if the function is monotone on some interval, as well as its derivative, the function doesn't have point(s) of inflection. In case the function is monotone, but its derivative is not, the function has point(s) of inflection on the interval.

When function is monotone, but its derivative is not, function changes type of its convexity still being monotone. Of course, this happens in the case if the necessary convexity condition is true.

Several people have already posted $x^3$ as example. Here are some plots: the function and its first and second derivatives.


Another example:

random function

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+1 for a general explanation instead of just a counter-example – Kevin Jan 14 '13 at 7:31

$f(x)=x^3$, with $-1<x< 1$.

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Just for another example $f(x)=x^3+x$ has non-zero derivative $3x^2+1$.

The only thing you need for $f(x)$ to be increasing is for the derivative to be non-negative (and if you want strictly increasing you need zeros of the derivative to be isolated). Take your favourite wiggly non-negative function (not too pathological) and integrate it - eg $\sin^2 x$.

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