Given a real matrix $F$, I'm trying to simplify the expression $$ C_{ijkl} = \frac{\partial (F_{ip}F_{jp})}{\partial F_{km}} F_{lm}$$
One reference I saw online says the right-hand side simplifies to: $$C_{ijkl} = \frac{1}{2} \left(\delta_{ik}F_{jp}F_{lp} + \delta_{jk}F_{ip}F_{lp} + \delta_{jl}F_{ip}F_{kp} + \delta_{il}F_{jp}F_{kp}\right)$$
Could someone please help me understand how the right-hand side would simplify to this? My first intuition was that we would evaluate the derivative by taking the product rule like below, but this gives me a different result: $$C_{ijkl} = (\delta_{ik}\delta_{pm}F_{jp} + \delta_{jk}\delta_{pm}F_{ip})F_{lm} = \delta_{ik}F_{jp}F_{lp} + \delta_{jk}F_{ip}F_{lp}$$
Thank you!
I agree with your result. They have probably used the fact that $C_{ijkl}$ is symmetric with respect to interchanges of $i$ and $j,$ and with respect to interchanges of $k$ and $l.$ The first term then equals the second, and the third equals the fourth. Splitting a tensor into symmetric and antisymmetric parts, and then noting that the product of the symmetric and antisymmetric parts is zero, is a common artifice in continuum mechanics.