Can we say anything about $\vert f'(x) \vert$ versus $\vert f''(x) \vert$ if $f$ is concave and goes through the origin? Suppose we have a function $f(x)$ that is concave, upward sloping, and goes through the origin.
Are we able to say anything about how $\vert f'(x)\vert$ compares to $\vert f''(x)\vert$, such as whether one is greater than/less than another?
 A: No. For any $a > 0$, $f(x) = 1-e^{-ax}$ satisfies those conditions, and $|f''(x)/f'(x)| = a$.
More general, let $h: \Bbb R \to \Bbb R$ be a continuous, positive function. Then
$$
 f(x) = \int_0^x e^{-\int_0^t h(s) ds} dt
$$
satisfies
$$
 f'(x) = e^{-\int_0^x h(s) ds} > 0 \\
 f''(x) = -h(x) e^{-\int_0^x h(s) ds} < 0
$$
so that $f$ is increasing and concave, with $f''(x)/f'(x) = -h(x)$, i.e. that ratio can be equal to an arbitrary prescribed negative continuous function.
A: Alternative approach.
Depending on your definition of concave, "Concave, upward sloping" is contradictory.  This answer assumes that you intend that $f(x)$ is convex everywhere.
Let $f(x) = ax^2 + bx \implies f$ passes through the origin.
$|f'(x)| = |2ax + b|~$ and $~|f''(x)| = |2a|.$
Convex merely implies 2nd derivative positive which implies that $a > 0$.
Consider the following three functions:
$f_1(x) = 1000x^2 + x~ \implies ~|f'_1(x)| = |2000x + 1|,~$
and $~|f''_1(x)| = 2000.$
For $~-(1/2) < x < (1/2), ~~|f'_1(x)| < |f''_1(x)|.$
For $~1 < x, ~~|f'_1(x)| > |f''_1(x)|.$

$f_2(x) = x^2 - x~ \implies ~|f'_2(x)| = |2x - 1|,~$
and $~|f''_2(x)| = 2.$
For $~0 < x < 1, ~~|f'_2(x)| < |f''_2(x)|.$
For $~3 < x, ~~|f'_2(x)| > |f''_2(x)|.$

$f_3(x) = x^2 + 1000x~ \implies ~|f'_3(x)| = |2x + 1000|,~$
and $~|f''_3(x)| = 2.$
For $~-(501) < x < -(500), ~~|f'_3(x)| < |f''_3(x)|.$
For $~0 < x, ~~|f'_3(x)| > |f''_3(x)|.$
