# How do you take the derivative with respect to a rank two tensor?

I am learning classical field theory and am trying to find the momentum density of the electromagnetic lagrangian as part of an example of Noether's Theorem. The derivative I am encountering is: $$\frac{\partial}{\partial(\partial_{\sigma}A_{\rho})} \left( \partial_{\mu}A_{\nu}\partial^{\mu}A^{\nu} \right).$$ I know that from here one uses chain rule to split it into two partial derivatives, and I know that in the end I should get a quantity that has upper indices $\sigma$ and $\rho$, since I am just taking the derivative of a scalar (indices are summed over).
I also realize that basically I am taking the derivative of "the square of the derivative of $A$", with respect to "the derivative of $A$", but I am confused as to how the indices work out.

edit: I should mention that all indices are 0,1,2,3.

• Every index combination indicates a different function. The derivative of a function w.r.t another functionally independent function, which is the case for functions with different indices, is zero and the only non-zero case is when the function to be derived and the one to be derived with are the same. – auxsvr Jun 20 '15 at 10:00
• Related: physics.stackexchange.com/q/3005/2451 and links therein. – Qmechanic Jun 20 '15 at 10:35

The answer: $$2\partial^\sigma A^\rho.$$ It's not the chain rule but the product rule you use. It is as if the tensor $$\partial_\mu A_\nu$$ were the variable you are differentiating with respect to. To understand the indices, consider a simpler example: $$\frac{\partial}{\partial x^i}(x^j x_j) = \frac{\partial}{\partial x^i}(x^j x^k g_{jk})= \delta^j_i x_j + \delta^k_i x_k = 2x_i$$.