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In some cases it turns out that,

If $f$ is strictly increasing, and convex then all its higher order derivatives are also strictly increasing and convex (assuming $f$ is continuously differentiable).

For example,

  • $e^{a x}$ on $\mathbb{R}$ for $a>0$,
  • $\tan(x)$ on $(0,\frac{\pi}{2})$ or,
  • $x^n$ on $(0,\infty)$ - at least for the first few derivatives.

Of course several counter examples, eg. $x \log x$, which is convex and increasing on $(1,\infty)$ but its derivatives do not share this property.

So my question is that under what conditions does it happen that if $f$ is increasing then its derivatives are also monotone (when are they increasing or decreasing)??

By conditions I mean conditions of $f$, not its higher derivatives. I know this is somewhat unusual in that I am trying to infer information about $f'$ knowing information about $f$. In most calculus courses the reverse is deduced.

Thanks for reading

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    Not all of the derivatives of $x^n$ are strictly increasing. Any idea what sort of conditions you're looking for? For what it's worth, if $f^{(n)}$ is increasing, then $f^{(n-1)}$ is convex, so your condition implies that all derivatives of $f$ are convex.2011-12-05
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    There's an obvious pattern the higher derivatives of the example functions possess. For example on the domain $(-\frac{\pi}{2},0)$, the first derivative of $tan(x)$ is decreasing, the 2nd derivative is increasing, the third decreasing and so on. Then on $(0,\frac{\pi}{2})$ they are all increasing. I want to know whats behind this pattern. I know of other functions (which are also increasing ans convex) who's derivatives follow similar patterns. So I thought maybe by knowing something about $f$ one could deduce that all its higher derivatives are increasing! Maybe its not the case tho!2011-12-05
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    In most calculus courses the reverse is deduced, but in most advanced analysis courses it is important to know what can we say about $f'$ if we know things about $f$ : convex functions have important differentiability properties, in particular that they are always *semi-differentiable*, i.e. differentiable in every direction (the limit when $h \to 0$ is only taken with $h \to 0^+$ though).2011-12-05

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