Partial derivative of a function with respect to a vector I have the following error term E:
$$E = \frac{1}{c}\sum_{\substack{
   i<j}} \frac{[d_{ij}^* - d_{ij}]^2}{d_{ij}^*}$$ 
where
$$c = \sum_{\substack{
   i<j}}d_{ij}^*$$
and 
$$d_{ij} = \sqrt{\sum_{\substack{k=1}}^{d} [ y_{ik} - y_{jk}]^2}$$
and
$$d^*_{ij} = \sqrt{\sum_{\substack{k=1}}^{n} [ y_{ik} - y_{jk}]^2}, n < d $$
$d_{ij}^*$ have the same equation as above but its values are constant but the measures are in $\mathbb{R}^{N}$ and $n < d$ . 
I need step by step explanation on finding the partial derivative $\frac{\partial E}{\partial y_{pq}}$ which is given below.
$$\frac{\partial E}{\partial y_{pq}} = \frac{-2}{c}\sum_{\substack{
   j=1,j\neq p}}^{N} \left[\frac{d_{pj}^* - d_{pj}}{d_{pj}d_{pj}^*}\right](y_{pq} - y_{jq}) $$
 A: I will give you some relevant steps of the computations.
We have
$$\frac{\partial E}{\partial y_{pq}}=-\frac{1}{c^2}\frac{\partial c}{\partial y_{pq}}\sum_{i<j}f(d_{ij},d^{*}_{ij})+ \frac{1}{c}\sum_{i<j}\frac{\partial f(d_{ij},d^{*}_{ij})}{\partial y_{pq}},$$
where $f(d_{ij},d^{*}_{ij})=\frac{d_{ij}-d^{*}_{ij}}{d^{*}_{ij}}$.
Now
$$\frac{\partial d_{ij} }{ \partial y_{pq} }=\frac{1}{d_{ij}} \sum_{k=1}^d\left( 2(y_{ik}-y_{jk})\frac{\partial (y_{ik}-y_{jk})}{\partial y_{pq}}\right),  $$
and
$$\frac{\partial d^{*}_{ij} }{ \partial y_{pq} }=\frac{1}{d^{*}_{ij}} \sum_{k=1}^n\left( 2(y_{ik}-y_{jk})\frac{\partial (y_{ik}-y_{jk})}{\partial y_{pq}}\right).  $$
Before computing $\frac{\partial (y_{ik}-y_{jk})}{\partial y_{pq}}$, using the above formulae we can arrive at
$$\frac{\partial c}{\partial y_{pq}}=\sum_{i<j}\frac{\partial d^{*}_{ij}}{\partial y_{pq}},$$
and
$$\frac{\partial f(d_{ij},d^{*}_{ij}) }{ \partial y_{pq} }=
\frac{\partial }{\partial y_{pq}}\left( \frac{ d_{ij}-d^{*}_{ij} }{ d^{*}_{ij} }\right)=
 -\frac{1}{(d^{*}_{ij})^2}\frac{\partial d^{*}_{ij} }{ \partial y_{pq} }(d_{ij}-d^{*}_{ij})+\frac{1}{d^{*}_{ij}}\frac{\partial (d_{ij}-d^{*}_{ij}) }{ \partial y_{pq} }= \\-\frac{1}{(d^{*}_{ij})^2}\frac{\partial d^{*}_{ij} }{ \partial y_{pq} }(d_{ij}-d^{*}_{ij})+\frac{1}{d^{*}_{ij}}\frac{\partial d_{ij} }{ \partial y_{pq} }  -\frac{1}{d^{*}_{ij}}\frac{\partial d^{*}_{ij} }{ \partial y_{pq} }.$$
All is left is the computation of $\frac{\partial (y_{ik}-y_{jk})}{\partial y_{pq}}$, with $i<j$. The idea is to use
$$\frac{\partial y_{ik}}{\partial y_{pq}}=\delta_{ip}\delta_{kq},$$
$$\frac{\partial y_{jk}}{\partial y_{pq}}=\delta_{jp}\delta_{kq}.$$
Now you need to sum all the above considerations and (after some work) arrive at the desired result.
