# [Math] Partial derivative and change of coordinates

multivariable-calculus

A colleague posted this on the door outside his office:

$$\frac{\partial}{\partial(x+y)}(xy)=?$$

Trying to be helpful, I gave it a shot:

$$u = x + y \\v = x – y \\ x = \frac{u+v}{2} \\ y = \frac{u-v}{2} \\ xy = \frac{u^2 – v^2}{4}$$

Then

$$\frac{\partial}{\partial(x+y)}(xy) = \frac{1}{4}\frac{\partial}{\partial u}(u^2 – v^2) = \frac{u}{2} = \frac{x+y}{2}.$$

Then when I got back to my desk I was concerned that I skimmed over some details and missed a constant somewhere, since I'm changing the scale with the change of variables. I'd like to try with

$$u' = \frac{x+y}{\sqrt{2}} \\ v' = \frac{x-y}{\sqrt{2}}$$

so that the scale doesn't change:

$$x = \frac{u' + v'}{\sqrt{2}} \\ y = \frac{u' – v'}{\sqrt{2}}$$

But then I'm left with interpreting

$$\frac{\partial}{\partial \left(\sqrt{2}u'\right)},$$

which I'm not sure how to do.

Does my conern matter?

The (correct) note by mfl gave me the hint I needed. I found a good explanation here.

If I have $u' = cu$ and $v' = cv$, then $du' = c\;du$ and $dv' = c\;dv$. (I have $u, v$ orthogonal, as well as $u', v'$.)

Then,

$$df = \frac{\partial f}{\partial u} du + \frac{\partial f}{\partial v} dv = \frac{1}{c}\frac{\partial f}{\partial u} du' + \frac{1}{c} \frac{\partial f}{\partial v} dv' = \frac{\partial f}{\partial u'} du' + \frac{\partial f}{\partial v'} dv',$$

from which follows

$$\frac{\partial}{\partial u'} = \frac{\partial}{\partial(cu)} = \frac{1}{c}\frac{\partial}{\partial u}.$$

Doing the problem with these variables doesn't change the answer.