= Solution
Work first with $|k|<1$, so the real integrands and their parameter <derivatives> are smooth on every finite amplitude interval. For the <elliptic integral of the second kind>,
$$
E_k=-k\int_0^\theta\frac{\sin^2x}{\sqrt{1-k^2\sin^2x}}dx.
$$
On the other hand, subtracting the integrands of the second-kind and first-kind <elliptic integrals> gives
$$
E-F=\int_0^\theta\frac{(1-k^2\sin^2x)-1}{\sqrt{1-k^2\sin^2x}}dx=-k^2\int_0^\theta\frac{\sin^2x}{\sqrt{1-k^2\sin^2x}}dx.
$$
Consequently
$$
\boxed{E_k=\frac{E-F}{k}\quad(k\ne0).}
$$
This is the <modulus derivative of the incomplete elliptic integral of the second kind>. At $k=0$ its right-hand side has a removable limit zero, agreeing with $E_k(\theta,0)=0$. Both $E$ and $F$ here use amplitude $\theta$ and modulus $k$; differentiating with respect to the alternative parameter $k^2$ would give a different prefactor.
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