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I have a function that is strictly monotonous on $[0,\infty)$ and $f(0)=0$. The question is whether the derivative of the function is convex.

My first thought was, there is not enough data to know that. But I realized I couldn't really think of a function that would be concave and satisfy upper conditions. My first thought was $\sqrt x$ but it is not derivable in $0$. Could I somehow extend it to make it derivable at 0? Is there any other function that would work?

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    please use tex to type your questions. Have a look at your question and to others questions via clicking on "edit" and see how you can type it correctly. About your question just take $y=x^2$, but clearly not all increasing funtions are convex or concave. Only a handful portion of them are convex.. – 2017-01-23
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    Wording tip: To say a function has a derivative at a point is to say it is differentiable there. – 2017-01-23
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    $\sqrt{x+1}$? $\sqrt{x+2}$? – 2017-01-24
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    @Michael that doesn't satisfy $f(0)=0$ – 2017-01-24
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    So you can just shift it down if you like: $-1+\sqrt{x+1}$. – 2017-01-24

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