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Say $f: \mathbb{R}^n \rightarrow \mathbb{R}$ by some $f(x_1,x_2,\cdots,x_n)$. Further suppose that each $x_i: \mathbb{R}^m \rightarrow \mathbb{R}$ by some $x_i=x_i(\eta_1,\eta_2,\cdots,\eta_m)$. Does it then follow that $$\frac{\partial f}{\partial \eta_i} = \sum_{k=1}^{n} \frac{\partial f}{\partial x_k}\frac{\partial x_k}{\partial \eta_i},$$ assuming all of $\frac{\partial f}{\partial x_k}$ and $\frac{\partial x_k}{\partial \eta_i}$ exist on some subset of $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively?

I don't have much formal education in way of partial derivatives and aside from the fact of treating variables that we aren't taking the derivative with respect to to be constants.

Really, my goal is to be able to successfully take some PDE like: $$u_{xx}+u_{yy}=0$$ with $u=u(x,y)$ and doing some coordinate change $\xi = x+y$ and $\eta=x-y$ and doing the change of variable to the homogenous equation. What do people call this? Doing a change of variable to the PDE or equation?

If this is correct I will be able to move forward in at least being able to do the chain rule I need to for my class.

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    $\begingroup$ Looks good. I like to think about the multivariable chain rule in the following way. How does $f$ vary with respect to $\eta_i$? Since $f$ depends on each $x_k$ and each $x_k$ depends on $\eta_i$, we can visualize a graph with a node for $\eta_i$ at the bottom, connected by an edge to each $x_k$ on a level above, and each of those nodes connected by an edge to $f$ above that. The edges represent partial derivatives, and the derivative $\frac{\partial f}{\partial \eta_i}$ is the sum of the products along each of the paths from $\eta_i$ to $f$. (There are $n$ of them in your example.) $\endgroup$ May 12 '13 at 23:50
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Yes; since you're taking only one $\eta_i$ to vary everything just reduces to the case where you just had $x_k(\eta)$ in fact.

One thing that is worth stressing is that the $\partial/\partial x_k$ derivatives are holding the other $x_m$ constant, rather than the other $\eta_j$ variables held constant elsewhere. This is probably implicit, but important to realize.

And I would just call it a change of variables!

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  • $\begingroup$ All right, now lets see if I can do this correctly ;) thanks $\endgroup$
    – user43138
    May 13 '13 at 1:37

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