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I want to compute the following derivative

$$ \frac{d}{d(x+y)^2}xy\,. $$

To do it I change variables $$ x+y\equiv \sqrt{2}u\quad\text{and}\quad x-y\equiv \sqrt{2}v\,, \qquad\text{so that}\qquad u+v= \sqrt{2}x\quad\text{and}\quad u-v=\sqrt{2}y\,, $$ to rewrite the derivative as $$ \frac{d}{d\left(\sqrt{2}u\right)^2}\frac{1}{2}\left(u^2-v^2\right)=\frac{1}{4}\,. $$

I have two questions:

  1. Is the derivative well defined?
  2. Does the procedure (and the result) make sense?
Ruth Murphy
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    No matter how you transform the variables, $xy$ is not a function of $(x+y)^2,.$ The answers are therefore no and no. – Kurt G. Feb 06 '24 at 12:40
  • Are you saying that the result of the derivative is zero or it is not even defined? In the second case why it is so? Intuitively if I change $x+y$ the function $xy$ changes in some way, right? – Ruth Murphy Feb 06 '24 at 13:40
  • A total derivative of the form $\frac{df}{dg}$ can be viewed as $$\frac{;\frac{df}{dx};}{;\frac{dg}{dx};}$$ but that's only defined when $f$ and $g$ are functions of $x$ only. – Kurt G. Feb 06 '24 at 13:48
  • Ah, maybe is there a way to generalize it? I found the following question, which seems to be coherent with my "solution": https://math.stackexchange.com/questions/307411/derivative-of-fx-y-with-respect-to-another-function-of-two-variables-kx-y – Ruth Murphy Feb 06 '24 at 13:57
  • Sorry, I should have checked for some answers before to post my question. I'll answer myself pointing at the previous link. Thanks anyway. – Ruth Murphy Feb 06 '24 at 14:02
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    A partial derivative needs to know what is being held constant at all times. If I had a proper coordinate change of $u=xy$ and $v=(x+y)^2$, then your derivative would be $\frac{\partial u}{\partial v}= 0$ for any pair of independent coordinates. On the other hand if I chose $u=x^2+y^2$, then your derivative would be $$\frac{\partial\left(\frac{v-u}{2}\right)}{\partial v} = \frac{1}{2}$$ You always need specify all of your variables, aka the full change of coordinates. – Ninad Munshi Feb 06 '24 at 19:14
  • The mapping between $(x,y)$ and $(u,v)$ should be invertible, no doubt that if you use these (non-invertible) mappings the derivative changes. – Ruth Murphy Feb 07 '24 at 16:45
  • @RuthMurphy your comment has no bearing on the point I was making. Of course all coordinate changes are invertible or have a domain of invertibility. The point is you do not have enough information to specify a partial derivative without a specifying a mapping from a full set of $n$ variables to another set of $n$ variables. There is no such thing as a canonical change of variables to enforce that unrigorous notation as you have attempted to do in the latest edit of your post. – Ninad Munshi Feb 16 '24 at 07:28

2 Answers2

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The derivative can be defined as the derivative of a function of two variables, $f(x,y)=xy$, with respect to another function of two variables, $g(x,y)=(x+y)^2$. The simple change of variables gives the correct answer, for a rigorous explanation check the following question: Derivative of $f(x,y)$ with respect to another function of two variables $k(x,y)$ .

Ruth Murphy
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The expression $$ \frac{d}{d(x+y)^2}xy $$ is undefined as a total derivative of a function $f(x,y)=xy$ with respect to another function $g(x,y)=(x+y)^2\,.$ The case when $$ \frac{df}{dg} $$ is defined is precisely when the two functions can be rewritten as functions of a single common variable $t\,.$ If this is the case can be tested with each of the following necessary conditions:

  • locally $f$ is a function of $g$ and vice versa

  • the two functions share the same level sets: $\{(x,y):f(x,y)=c\}=\{(x,y):g(x,y)=c'\}$

  • $df\wedge dg\equiv 0\,.$ See this for a proof.

Plotting $xy$ and $(x+y)^2$ in GeoGebra or Desmos should immediately convince you that they don't have the same level sets.

Also: $df=x\,dy+y\,dx$ and $dg=2(x+y)(dx+dy)$ so that $$ df\wedge dg=2x(x+y)\,dy\wedge dx+2y(x+y)\,dx\wedge dy=2(y^2-x^2)\,dx\wedge dy $$ which does not vanish identically.

Kurt G.
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