Transformation that makes derivative of a function non-negative everywhere Is there an analytical and differentiable transformation $g$ that ensures that $d g(f(x)) / dx$ is non-negative everywhere and that does not changes the magnitude of derivative too much, or that at least preserves ordering, i.e.:
if
$$\frac{df(x_1)}{dx} > \frac{df(x_2)}{dx}$$
then
$$\frac{d g(f(x_1))}{dx} > \frac{d g(f(x_2))}{dx}?$$
Additionally, if there is no $g$ transformation for a "general" $f$, consider if there is one for the case where $f$ is a neural network (linear operations on $x$ followed by non-linear activation functions such as ReLU and additional linear operations and activations).
 A: An example intended to induce you to clarify your use of "analytical transformation":
$$  g(f)(x) = \int_0^x \frac{\pi}{2} + \arctan\left( \frac{\mathrm{d}}{\mathrm{d}t}f(t) \right) \,\mathrm{d}t  $$
This $g$ is analytic in the sense that it agrees with its power series on a neighborhood of $f \equiv 0$.
As an example, for this $g$ and $f(x) = x^2$, $$g(f)(x) = \frac{\pi x}{2} + x \arctan(2x) - \frac{1}{4} \ln(4x^2 + 1)  $$

having
$$ \frac{\mathrm{d}g(f)(x)}{\mathrm{d}x} = \frac{\pi}{2} + \arctan{2x}  \text{.}  $$
Many others like this can be made along the recipe: pick a monotonically increasing function, $h$ on $\mathbb{R}$ with a lower horizontal asymptote (examples: $\arctan x$, $\mathrm{e}^x$, any sigmoidal function).  Then let $m$ be the height of that lower horizontal asymptote and let $x_0$ be the $x$-intercept of $g(f)$.  Then 
$$  g(f)(x) = \int_{x_0}^x m + h\left( \frac{\mathrm{d}}{\mathrm{d}t} f(t) \right) \,\mathrm{d}t $$
is such a $g$.
Picking $h(x) = \mathrm{e}^x$, $m = 0$, $x_0 = -1$ and repeating $f(x) = x^2$, we get 
$$  g(f)(x) = \frac{\mathrm{e}^{2x+2} - 1}{2\mathrm{e}^2}  $$

having
$$ \frac{\mathrm{d}g(f)(x)}{\mathrm{d}x} = \mathrm{e}^{2x}  \text{.}  $$
Note that this recipe essentially slavishly enforces your monotonicity of derivatives and then positivity of the derivatives of the result by grabbing the derivatives, applying a monotonically increasing function to them, shifting them up so that the minimum possible of the transformed derivatives is zero, then integrating.
A: Partial Answer
In a grand abuse of notation, let me show you how I arrived at one possibility. Note that
$$\frac{dg(f(x))}{dx}=\frac{dg}{df}\bigg|_{f(x)}\cdot\frac{df}{dx}, $$
by the chain rule. So one solution would be if we could make
$$\frac{dg}{df}\bigg|_{f(x)}=\frac{df}{dx}, $$
making
$$\frac{dg(f(x))}{dx}=\left(\frac{df}{dx}\right)^{\!2}\ge 0. $$
So, let us compute formally:
\begin{align*}
\frac{dg}{df}\bigg|_{f(x)}&=\frac{df}{dx} \\
\int\frac{dg}{df}\bigg|_{f(x)}\,df(x)&=\int\frac{df}{dx}\,df(x) \\
g(f(x))&=f(x)\,f'(x).
\end{align*}
Double-check:
$$\frac{dg}{dx}=\frac{dg}{df}\,\frac{df}{dx}=f'(x)\,f'(x)=(f'(x))^2, $$
which works. Note that, in these computations, $f$ is independent of $f',$ each considered as a variable. This is standard practice in the Calculus of Variations, so there is precedent.
I rather fancy this preserves the ordering, as well.
[EDIT] This does not preserve the ordering, as Eric Towers has pointed out in his comment. However, this idea could be a spring-board for another solution, so I will not delete.
