Partial derivative with respect to metric tensor $\frac{\partial}{\partial{g_{kn}}}(g_{pj}g_{ql})$ $$-\frac{1}{4\mu_0}F^{pq}F^{jl} \frac{\partial}{\partial{g_{kn}}}(g_{pj}g_{ql})=+\frac{1}{4\mu_0} F^{pq} F^{lj} 2 \delta^k_p \delta^n_j g_{ql}$$
I need to know how to derive $\frac{\partial}{\partial{g_{kn}}}(g_{pj}g_{ql})$. Can you explain how the final result on the right side was obtained?
 A: This is a consequence of the symmetry of the metric tensor (which is what I'm assuming $g_{pq}$ is). 
Clearly,
$$
\frac{\partial}{\partial g_{kn}} (g_{pj}g_{ql}) = \frac{\partial g_{pj}}{ \partial g_{kn}} g_{ql} + g_{pj} \frac{\partial g_{ql}}{\partial g_{kn}} 
$$
The partial derivative in this expression must give something symmetric. For a non-symmetric tensor $a$, all 9 components are independent and $\dfrac{\partial a_{pq}}{\partial a_{kn}}$ is non-zero only when $p=k$ and $q=n$, i.e. you have
$$
\dfrac{\partial a_{pq}}{\partial a_{kn}} = \delta_{p}^{k} \delta_{q}^{n}
$$
For a symmetric tensor, $g_{ij}=g_{ji}$ so you can't do that. Instead, you must symmetrize the argument by rewriting it as 
$$g_{pj} = \frac{1}{2} (a_{pj} + a_{jp})
$$
Now take partial derivatives using the formula derived above for the non-symmetric tensor.
You'll get
$$\frac{\partial g_{pj}}{\partial g_{kn}} = \frac{1}{2} (\delta_{p}^{k} \delta_{j}^{n} + \delta_{j}^{k} \delta_{p}^{n}) \\
$$
You can proceed similarly for the other term $\dfrac{\partial g_{ql}}{\partial g_{kn}}$.
Finally you have
$$
\frac{\partial}{\partial g_{kn}} (g_{pj}g_{ql}) = \frac{1}{2} (\delta_{p}^{k} \delta_{j}^{n} + \delta_{j}^{k} \delta_{p}^{n}) g_{ql} + \frac{1}{2} (\delta_{q}^{k} \delta_{l}^{n} + \delta_{l}^{k} \delta_{q}^{n}) g_{pj} 
$$
Now consider the contraction of this expression with $F^{pq} F^{jl}$ and proceed. 
