I have come to realize another very practical reason for teaching differentiation before integration.
In most applications of integration we are splitting something (area, volume, arclength, work, etc) into lots of tiny pieces, compute an approximation of each piece, and then sum. Taking the limit as the size of the pieces tends to $0$ yields an integral.
What is underappreciated, I think, is that often the "approximate each piece" step will involve differentiation.
Take arc length as an example.
We have a function $f: [a,b] \to \mathbb{R}$ and we want to calculate the arc length of the graph of $f$.
It is natural to first subdivide the interval $[a,b]$ into $N$ equal sized subintervals $[x_k, x_{k+1}]$ with $x_k = a + \frac{b-a}{N}(k-1)$ for $k=0,1,2, 3 \dots, N, N+1$.
Since arclength is additive, the length of the whole curve is the sum of the length over each subinterval.
We then approximate each small arc by the length of the secant line connecting the two endpoints $\sqrt{(f(x_{k+1}) - f(x_k))^2 + (x_{k+1}-x_k)^2}$.
The issue is that
$$\sum_0^{N-1} \sqrt{(f(x_{k+1}) - f(x_k))^2 + (x_{k+1}-x_k)^2}$$
is not in the form of a Riemann sum for any function.
So we need to approximate each of these summands again using the derivative:
$$
\sqrt{(f(x_{k+1}) - f(x_k))^2 + (x_{k+1}-x_k)^2} \approx \sqrt{(f'(x_k)(x_{k+1}-x_k))^2 + (x_{k+1}-x_k)^2}
$$
So we obtain
$$\sum_0^{N-1} \sqrt{(f'(x_k))^2 + 1} \Delta x$$
as our approximation of the arc length.
This is an extremely common phenomenon: forming an integral to compute a quantity of interest relies on decomposition into small pieces, but approximating these small pieces requires differentiation. This is a good mathematical reason for studying differentiation first.