# $Q=\Sigma q_i$ and its differentiation by one of its variables

Suppose that $Q = q_1 + ... +q_n$.

why is $$\frac{dQ}{dq_i} = \Sigma_{j=1}^{n}\frac{\partial q_j}{\partial q_i}$$?

Is it related to each $q_i$ being independent to other $q_i$'s?

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No, it is because differentiation is linear, ie, if $h=f+g$, then $\frac{d h}{d x} = \frac{d f}{d x} + \frac{d g}{d x}$.
so suppose that each $q_i$ is independent variable - does this mean that $dq_i/dq_j$ and $\partial q_i/\partial q_j$ are zero whenever $i \neq j$? – Lanz Feb 2 '13 at 6:23
Perhaps you should elaborate the question slightly, and show the arguments to $Q$ and $q_j$, I may have misinterpreted the question... – copper.hat Feb 2 '13 at 6:25
I didn't read your question carefully. I interpreted your questions as something like $Q(x_1,...) = q_1(x_1,...)+...+q_n(x_1,...)$ and then $\frac{\partial Q(x_1,...)}{\partial x_i} = \sum_k \frac{\partial q_k(x_1,...)}{\partial x_i}$. I don't know how to interpret $\frac{\partial q_j}{\partial q_i}$. – copper.hat Feb 2 '13 at 6:33