If $f$ is strictly increasing and convex then what about $f'$? 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
 A: One property that will give us $f'$ strictly increasing is the following : we say that $f$ is convex if
$$
\forall \lambda \in [0,1], \quad f(x + \lambda( y-x)) \le f(x) + \lambda( f(y) - f(x) )
$$
and we say that $f$ is strictly convex if the inequality is always strict. (The intuition behind this is that if you draw the cord between the points $(x,f(x))$ and $(y,f(y))$, the cord cannot touch the graph, thus we cannot have an interval where the derivative is constant if $f$ is strictly increasing.) 
Assume $f$ is strictly convex and differentiable. Then $f'$ is strictly increasing. Since strictly convex functions are convex, we already know that $f'$ is increasing. Suppose there exists $x \le y$ such that $f'(x) = f'(y)$. Then this means that in the interval $[x,y]$, the function $f'$ is constant. But then $f$ cannot be strictly increasing on this interval unless $x=y$, thus $f'$ must be strictly increasing. The reason for this is that strict convexity on $[x,y]$ with $x < y$ implies that by the mean value theorem, there exists $c_1$, $c_2 \in ]x,y[$ with the property that
$$
\begin{align}
f(x + \lambda (y-x)) - f(x) & < \lambda (f(y) - f(x)) \\
\\
\Longrightarrow \quad f'(c_1) = \frac{ f(x + \lambda(y-x)) - f(x)}{\lambda(y-x)} & < \frac{f(y) - f(x)}{y-x} = f'(c_2)
\end{align}
$$
which is a contradiction. One way to ensure that a function has a strictly decreasing derivative is strict concavity : a function is strictly concave if equality is reversed in the definition of strict convexity and the proof is similar. Is this what you were looking for? I hope it helps.
A: Let $V$ be the vector space of solutions to a linear differential equation with constant coefficients.  Then differentiation is a linear operator $D:V\to V$.
Suppose that $f\in V$ is an eigenvector of $D$ with positive eigenvalue.  Then, clearly, if $f$ is increasing and convex, so are $Df, D^2f,\ldots$
This is why $e^{ax}$ has this property; it is the (eigen) solution to the differential equation $Df = af$.
The reason that $x^n$ almost has this property is that it is almost an eigenvector.  Specifically, in the space of solutions to the equation $D^{n+1} f = 0$, $x^n$ is a generalized eigenvector of $D$.  Specifically, its derivatives give a basis for $V$: $\{x^n,nx^{n-1},\ldots\}$.  Note that each entry in this basis is strictly increasing, and convex on $(0,\infty)$.  And so positive combinations of them will be also.
The situation with $\tan x$ is a little more complicated, but not much.  Both $\tan x$ and $\sec x$ are increasing on $(0,\pi/2)$, and every derivative of $\tan x$ is a polynomial in $\tan x$ and $\sec x$ with positive coefficients.  So each time we differentiate, we're remaining in a space of functions which are increasing and convex.
So the answer to this question appears to be: if $f_1,\ldots , f_n$ are increasing, convex functions, and for all $i$, $Df_i$ is a non-zero polynomial in $f_1,\ldots , f_n$ with positive coefficients, then, for any non-zero polynomial $g$ in $f_1,\ldots , f_n$ with positive coefficients, every derivative of $g$ is increasing and convex.  In other words, I interpret this as a question as having more to do with algebra than analysis.
