calculus - $f$ convex strictly decreasing function , is $f'(x+delta)-f'(x)$ convex

Assume you have a strictly decreasing convex differentiable function $f(x)$, $x \in \Bbb R^+$, I am wondering if the increment of the first derivative is also convex; i.e.,

$$g(x) = f'(x+\delta) - f'(x)$$ where $\delta$ is any positive number.

What I concluded :

I can say that $f'(x)$ is a strictly increasing function, also since $f(x)$ is strictly decreasing, $f'(x)$ is always negative, meaning that it increases and approaches zero as $x \to \infty $, now I can visualize $f'$ as concave and the difference : $f'(x+\delta)-f'(x)$ to be decreasing but not sure how to show its convexity (if it is).

Answer

To complete gerw's comment and to avoid leaving the question unanswered, let me give an explicit counterexample.

The question (letting $g=f'$) is equivalent to the following one: If $g\colon(0,\infty)\to\mathbb{R}$ is continuous, strictly increasing and bounded above, is $g(x+\delta)-g(x)$ convex for $\delta>0$? Here is a counterexample:
$$

g(x)=\frac{x-\sin x}{1+x-\sin x}.
$$
We have
$$
g'(x)=\frac{1-\cos x}{(1+x-\sin x)^2}\ge0,\quad 0\le g(x)\le1.
$$
$g$ is increasing, bounded above and $\lim_{x\to\infty}g(x)=1$, but it is not concave. Here is the graph of $g(x+\delta)-g(x)$ for $\delta=0,1+0,2\,k$, $k=0,1,2,3,4$.

Clearly $g(x+\delta)-g(\delta)$ is not convex.

If you want a counterexample in terms of $f$ let
$$
f(x)=\int_0^x(g(t)-1)\,dt.
$$