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I was looking through a paper that described a simple payoff function where there is an outcome variable $Y$ that depends on some causation variable $X$ and the payoff is given as some function of the two as $\pi(X,Y)$. Now, when they took the derivative w.r.t to $X$, they ended up with this equation:

$\dfrac{d\pi(X,Y)}{dX} = \dfrac{\partial\pi}{\partial{X}}(Y) + \dfrac{\partial\pi}{\partial{Y}}\dfrac{\partial{Y}}{\partial{X}}$

I don't understand how this equation makes sense for a generic two-variable function. I suppose that since $Y$ depends on $X$ in some (unknown) way, this problem could be simplified as $Y=f(X)$ but even then, this doesn't seem to hold for some example payoff functions. For example, if $\pi(X,Y) = X+Y$, the $Y$ term in the derivative equation would be removed.

I feel like I'm missing something obvious here, just not sure what it is.

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  • $\begingroup$ Duplicate of this question? $\endgroup$ – amd Sep 21 '15 at 7:07
  • $\begingroup$ Not quite. Its similar, but in this case, $Y$ multiplying with the first term doesn't quite make sense. For some clarity, the author mentioned that the $Y$ multiplier is the part of the derivative relating to prediction and the $\frac{\partial{Y}}{\partial{X}}$ was the part relating to causation in economic analysis. $\endgroup$ – user3425451 Sep 21 '15 at 9:32
  • $\begingroup$ I understood that term as $\frac{\partial\pi}{\partial X}$ evaluated at $Y$. If that’s multiplication by $Y$ instead, we’ll need more context. $\endgroup$ – amd Sep 21 '15 at 10:04
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By the chain rule, $\frac{d\pi(X,Y)}{dX}=\frac{\partial\pi}{\partial X}\frac{\partial X}{\partial X}+\frac{\partial\pi}{\partial Y}\frac{\partial Y}{\partial X}.$ $\frac{\partial X}{\partial X}$ is, of course, $1$, leading to the r.h.s. in the formula you cite. The author appears to be emphasizing that the $\pi_X$ term is a function of $Y$.

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  • $\begingroup$ The main point I'm confused about is the $Y$ that is multiplying in the first term. In the example you're describing, you've omitted $Y$ in that term which is what I would have expected. Maybe the author was implicitly assuming some special type of function like $\pi(X,Y) = XY$ $\endgroup$ – user3425451 Sep 21 '15 at 9:30
  • $\begingroup$ That particular function wouldn’t give you a $Y$ times $\frac{\partial\pi}{\partial X}$ , though: $\frac{d(XY)}{dX}=Y+X\frac{dY}{dX}$. $\endgroup$ – amd Sep 21 '15 at 9:55
  • $\begingroup$ Ok, that would make a lot more sense. I guess I misinterpreted the notation. Thanks. $\endgroup$ – user3425451 Sep 21 '15 at 15:26

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