It can be easily checked that both the complete elliptic integrals $K(k), K'(k)$ satisfy the same second order differential equation $$kk'^{2}\frac{d^{2}y}{dk^{2}} + (1 - 3k^{2})\frac{dy}{dk} - ky = 0$$ and hence from the theory of second order differential equations there is a relation of the form $$K'(k) = cK(k)\cdot\log k + f(k)$$ where $c$ is some constant and $f(k)$ is some analytic function of $k$. The exact relation between $K(k)$ and $K'(k)$ is given by $$K'(k) = \frac{2K(k)}{\pi}\log\left(\frac{4}{k}\right) - 2\left[\left(\frac{1}{2}\right)^{2}\left(\frac{1}{1\cdot 2}\right)k^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{2}\left(\frac{1}{1\cdot 2} + \frac{1}{3\cdot 4}\right)k^{4} + \cdots\right]$$ It can be verified with some patience that the RHS does satisfy the differential equation given above and thereby the relation between $K'(k)$ and $K(k)$ can be established.
However is there an alternative proof based on the definition of $K'(k)$ and $K(k)$ as complete elliptic integrals or using the hypergeometric relation $$\frac{2K(k)}{\pi} =\,_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; k^{2}\right)$$ which can be presented to someone unaware of the theory of second order differential equations?
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