"Differential equation" is a misnomer here I'd say, since what you're trying to do is just maximise productivity here; hence you're solving an optimisation problem.

I'm not too sure why the \( \frac{100^2}{x^2} \) term is meant to represent here, but I'll trust that you understand your design well enough. The \(\sqrt{2x^2} \) part makes sense to me though.

So, to help combat the actual differentiation, here are my suggestions:

1. Firstly, because you're just assuming that \(x\) is a distance here, you should be safe to assume that \(x\geq 0\). Hence the \(\sqrt{2x^2}\) term can be simplified to \(\sqrt{2}x\). (Here, the \(\sqrt{2}\) has now basically become the coefficient of that \(x\) term.)

2. If you don't mind using the quotient rule a little, you may consider trying to simplify \(1 - \frac{1}{0.0019(x+6.6)^3} \) into one common denominator. Same goes for \(1 - \frac{1}{0.0019(\sqrt{2}x+6.6)^3} \).

3. Since the two powers happen to coincide, you may wish to consider merging them under the common power into this:

\[ \left( \left(1 - \frac{1}{0.0019(x+6.6)^3}\right)\left(1 - \frac{1}{0.0019(\sqrt2 x+6.6)^3}\right) \right)^4 . \]

From here, you could just expand the brackets inside the root. Alternatively, you could combine this wth point 2, and expand after combining the fractions.

4. The \( \frac{1}{x^2} \) term can technically also be moved in under the power, if you rewrite it as \( \left(\frac{1}{\sqrt{x}}\right)^4 \). Personally not a fan of this suggestion though.

5. You may instead wish to just use the generalised product rule to help you out. The version for three terms is mentioned on the Wikipedia page for the product rule.

6. This stretches the boundaries of maths methods a lot, because technically speaking it requires something from spesh. But if you're willing to go the extra mile in research (and if your assignment lets you do so...), you may wish to consider researching a technique called **logarithmic differentiation**. It's extra time trying to learn the mechanics of it, but it can help reduce a lot of the product rule and chain rule computations required.

**In saying this:** I have a bad feeling that you may be unable to solve for \(x\), when you set \( \frac{dp}{dx} = 0 \) (where \(p = productivity\)). I could be assuming things too quickly, but for all I know, it may be a problem that falls outside the limitations of basic algebra.

Of course, you may also wish to consider a simpler design. But that's not something I would recommend for, nor recommend against.