This question already has an answer here:

Assume that $F[h(\xi);x,y]$ be the inverse of $G[h(\xi);x,y]$ in the sense that the following identity is satisfied: \begin{equation} \int dz F[h(\xi);x,z]G[h(\xi);z,y] \equiv \delta(x-y) \end{equation} for any $h(\xi)$. Then one can obtain: \begin{equation} \dfrac{\delta}{\delta{h(z)}}F[h(\zeta);x,y] = -\int d\xi d\eta F[h(\zeta);x,\xi]\dfrac{\delta G[h(\zeta);\xi,\eta]}{\delta{h(z)}}F[h(\zeta);\eta,y] \end{equation} Question:How can I prove/understand this functional derivative?

Note: This functional derivative appears in this paper. [APPENDIX A:(A.7)]


marked as duplicate by Qmechanic May 22 at 18:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ This is done here. $\endgroup$ – Sunyam May 22 at 14:36
  • $\begingroup$ Thanks for your help. $\endgroup$ – Jack May 23 at 0:35