# Must a monotone function have a monotone derivative?

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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+1 just because it's so funny. – Git Gud Jan 13 at 20:23
It's 2-valued at one interval. Need eraser. – alancalvitti Jan 13 at 20:24
We need more pictures on MSE :) – Stuart Jan 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 at 1:16

Strict mathematical proof can be based on the convexity of the function, which depends on the second derivative of the function.

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

When function is monotone, but its derivative is not, function change 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 plots of this function, its first and second derivatives. I can't post images, so here is a link to the pictures I've made to help clear this out:

Random function

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 +1 for a general explanation instead of just a counter-example – Kevin Jan 14 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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