''' I have problem to drive the below derivation from double sigma equation please help me. The picture is here
I have problem to drive the below derivation from double sigma equation please help me.
$$W=\frac12\cdot\sum_{i=1}^6 \sum_{j=1}^6 C_{ij} \cdot \sigma_i\cdot \sigma_j$$
$$\frac{\partial W}{\partial \sigma_i \partial \sigma_j}=C_{ij}$$
$C_{ij}=$constant
It seems you want to differentiate W twice. WLOG, I take $m=n=3$ instead of $m=n=6$
$$W=\frac12\cdot \left(\sum_{i=1}^3 \sum_{j=1}^3 C_{ij} \cdot \sigma_i\cdot \sigma_j\right)$$
The sum can be written without the sigma signs.
$$W= \frac12\cdot \left( C_{11} \cdot \underbrace{\sigma_1\cdot \sigma_1}_{=\sigma_1^2}+C_{12} \cdot \sigma_1\cdot \sigma_2+\ldots +C_{21} \cdot \sigma_2\cdot \sigma_1+\ldots +C_{3} \cdot \sigma_3\cdot \sigma_3\right)$$
Now we can differentiate it w.r.t $\sigma_1$
Only the summands with at least one $\sigma_1$ is not $0$
$$\frac{\partial W}{\partial \sigma_1}=\frac12\cdot \left(C_{11}\cdot 2\sigma_1+C_{12}\cdot \sigma_2+C_{13}\cdot \sigma_3+C_{21}\cdot \sigma_2+C_{31}\cdot \sigma_3\right)$$
At the next step it can be differntiated w.r.t $\sigma_2$. The summands without a factor $\sigma_2$ become $0$
$$\frac{\partial W}{\partial \sigma_1 \partial \sigma_2}=\frac12\cdot \left(C_{12}+C_{21}\right)$$
The result can be generalized as
$$ \frac{\partial W}{\partial \sigma_i \partial \sigma_j}=\frac{C_{ij}+C_{ji}}2$$
If $C_{ij}=C_{ji}$ it becomes $\frac{C_{ij}+C_{ij}}2=\frac{2\cdot C_{ij}}2=C_{ij}$