The derivate formula in Matrix Cookbook As said in the book, The basic assumptions about matrix derivates can be written in a formula as 
$
\frac{∂X_{kl}}{∂X_{ij}} = δ_{ik}δ_{lj}
$
But I don't know how I can use this formula to calculate matrix derivates. Could anyone give some examples?
 A: A more intuitive description can be seen if you draw analogy with a standard derivative, of a function $f(x,y) = \alpha\beta$,
$$
\frac{\partial}{\partial x\partial y} \alpha\beta = 1 \qquad \text{if}\ \alpha\beta = xy\\
\frac{\partial}{\partial x\partial y} \alpha\beta = 0 \qquad \text{if}\ \alpha\beta \ne xy
$$
You can combine this by using the kroneka-delta as,
$$
\frac{\partial}{\partial x\partial y} \alpha\beta = \delta_{\alpha x}\delta_{\beta y}
$$
I then take this following example from Physics as demonstration. Consider the following (Lagrangian) where we have implied Einstein summation across indices.
$$
\mathcal{L} =
\tfrac{1}{2}
\left(\partial_{\mu} \varphi_{\nu}\right)
\left(\partial_{\mu} \varphi_{\nu}\right)
-
\tfrac{1}{2}m^2\varphi_\mu\varphi_\mu
$$
If we take a derivative with respect to $\left(\partial_\alpha\varphi_\sigma\right)$, as in the Equations of Motion, which can be put into the following form, $\left(\partial\varphi\right)_{\alpha\sigma}$ to be consistent with your question. Then the derivative explicitly,
$$
\frac{\partial\mathcal{L}}
{\partial\left(\partial_\alpha\varphi_\sigma\right)}
=
\left(\partial_{\mu} \varphi_{\nu}\right)
\left(
\frac{\partial}{\partial\left(\partial_\alpha\varphi_\sigma\right)}
\partial_{\mu} \varphi_{\nu}\right)
+
\left(
\frac{\partial}{\partial\left(\partial_\alpha\varphi_\sigma\right)}
\partial_{\mu} \varphi_{\nu}\right)
\left(\partial_{\mu} \varphi_{\nu}\right)
$$
Then just focusing on the derivative itself,
$$
\frac{\partial}{\partial\left(\partial_\alpha\varphi_\sigma\right)}
\partial_{\mu} \varphi_{\nu}
=
\delta_{\alpha\mu}\delta_{\sigma\nu}
$$
We can see how this interacts with the remaining term that was left untouched as part of the product rule,
$$
\left(
\frac{\partial}{\partial\left(\partial_\alpha\varphi_\sigma\right)}
\partial_{\mu} \varphi_{\nu}\right)
\left(\partial_{\mu} \varphi_{\nu}\right)
=
\delta_{\alpha\mu}\delta_{\sigma\nu}
\partial_{\mu} \varphi_{\nu}
=
\partial_{\alpha} \varphi_{\sigma}
$$
As a side note then the full derivative is then,
$$
\frac{\partial\mathcal{L}}
{\partial\left(\partial_\alpha\varphi_\sigma\right)}
=
\partial_{\alpha} \varphi_{\sigma}
$$
Note: I have taken the metric to be Euclidean so $g^{\mu\nu}=1$ on the diagonal else 0. If you want to use a Minkowski metric such as $(1,-1,-1,-1)$ then you have to be more careful.
