Differentiation of tensor product I have a tensor equation 
$$\frac{\partial A_{ij}}{\partial B_{kl}}=\frac{\partial A_{ij}}{\partial C_{pq}}\frac{\partial C_{pq}}{\partial B_{kl}}
$$
$C_{pq}$ can be written as $C_{pq}=B_{pq}+aB_{mm}\delta _{pq}$, where $a$ is scalar.
So I get,
$$\frac{\partial A_{ij}}{\partial B_{kl}}=\frac{\partial A_{ij}}{\partial C_{pq}} \bigg( \frac{\partial B_{pq}}{\partial B_{kl}}+a\frac{\partial (B_{mm} \delta_{pq})}{\partial B_{kl}}  \bigg)$$
Question 1: What is result of $$\frac{\partial B_{pq}}{\partial B_{kl}} \quad \textrm{and} \quad \frac{\partial (B_{mm} \delta_{pq})}{\partial B_{kl}}$$ 
Looks like it is 
$$\frac{\partial B_{pq}}{\partial B_{kl}} = \frac{1}{2}(\delta _{pk} \delta _{ql} + \delta _{pl} \delta _{qk}) $$
and 
$$
\frac{\partial (B_{mm} \delta_{pq})}{\partial B_{kl}}= \delta _{pq} \delta _{kl}
$$ 
Can someone comment whether above 2 results are correct and how to prove them?
After some simplification, 
$$\frac{\partial A_{ij}}{\partial C_{pq}}= \delta _{ip} \delta _{jq}$$
(This result is got after putting values of tensors and not by itself.)
Question 2: So, based on results above, is the following product correct ?
$$(\delta _{ip} \delta _{jq})(\delta _{pq} \delta _{kl})=\delta _{iq} \delta _{jq} \delta _{kl}$$
The left hand side has $i,j,k,l$ as free indices and above equation also has them with $q$ summed over index.
Thanks in advance.
 A: 
Looks like it is  $$\frac{\partial B_{pq}}{\partial B_{kl}} =
\frac{1}{2}(\delta _{pk} \delta _{ql} + \delta _{pl} \delta _{qk}) $$ and 
  $$ \frac{\partial (B_{mm} \delta_{pq})}{\partial B_{kl}}= \delta _{pq} \delta _{kl} $$

Partial derivatives are only well-defined when you specify what is held constant. We usually adopt a convention obtaining $$\frac{\partial B_{pq}}{\partial B_{kl}}=\delta_{pk}\delta_{ql}$$for general rank-$2$ tensors and $$\frac{\partial B_{pq}}{\partial B_{kl}}=\frac12(\delta_{pk}\delta_{ql}+\delta_{pl}\delta_{qk})$$for symmetric rank-$2$ tensors. In either case, your second result follows by contracting the repeated index $m$.

based on results above, is the following product correct ?
  $$(\delta _{ip} \delta _{jq})(\delta _{pq} \delta _{kl})=\delta _{iq}\delta _{jq} \delta _{kl}$$
  The left hand side has $i,j,k,l$ as free indices and above equation
  also has them with $q$ summed over index.

That follows if you contract $p$. I assume you're using it in $$\frac{\partial A_{ij}}{\partial B_{kl}}=\delta_{ip}\delta_{jq}\left(\frac{\delta_{pk}\delta_{ql}+\delta_{pl}\delta_{qk}}{2}+a\delta_{pq}\delta_{kl}\right)=\frac{\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}}{2}+a\delta_{ij}\delta_{kl}.$$
