I am new to elliptic integrals and did not learn about them in a formal setting (I learned about them from Wikipedia and my own fiddling). I recently attempted to take the derivative of the complete elliptic integral of the first kind with respect to $k^2$ such that I was performing $$\frac{\partial K(\phi)}{\partial \phi} = \frac{\partial}{\partial \phi} \int_0^\frac{\pi}{2}\frac{d\theta}{\sqrt{1 - \phi \sin^2\theta}}$$ where $\phi = k^2$. In so doing, I arrived at the very same road block asked about in this question. The answer to this question gives a proof that the derivative can be expressed in terms of elliptic integrals of the first and second kind (which is amazing to me and I consider quite lucky). The approach of the proof was surprising and seemed to come out of nowhere. I consider myself a moderately skilled mathematician and I would never have been able to come up with such a solution, especially due to the fact that, while those integrals in the question are in fact equal, their integrands are not.
My question is very simple. Is this derivative considered a mathematical discovery that took considerable time and thought, or is it considered to be obvious? Should I feel bad that this was nowhere near obvious to me?
P.S. I was able to evaluate the integral above using the result of $\frac{\partial K(k)}{\partial k}$ from the referenced question and the fact that $$\frac{\partial K(k)}{\partial k}\Bigg\rvert_{k = \sqrt{\phi}} = \frac{\partial K(\phi)}{\partial \phi}\frac{\partial \phi}{\partial k}\Bigg\rvert_{k = \sqrt{\phi}}$$ thus $$\frac{\partial K(\phi)}{\partial \phi} = \frac{1}{2\phi}\left(\frac{E(\phi)}{1 - \phi} - K(\phi)\right)$$
