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The question is related to this one.

Let $f(x,y) = xy$ and $g(x,y) = f(x,y)^2 = x^2y^2$. Now consider the fraction of partial derivatives \begin{align} \frac{\frac{\partial g}{\partial x}}{\frac{\partial f}{\partial x}} = \frac{2xy^2}{y} = 2xy \end{align}

I was wondering if I can cancel $\partial x$ in the fraction, i.e., \begin{align} \frac{\frac{\partial g}{\partial x}}{\frac{\partial f}{\partial x}} = \frac{\partial g}{\partial f} = \frac{\partial f^2}{\partial f} = 2f = 2xy. \end{align}

  • Is this some general principle or coincidence?
clueless
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  • Someone calls it chain rule – N74 Sep 28 '16 at 21:22
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    Here is another example. I mean you cannot cancel the partial derivative sign in principle can you? – clueless Sep 28 '16 at 21:26
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    The chain rule says that $\frac{\partial g}{\partial x}={\frac{\partial f}{\partial x}} \frac{\partial g}{\partial f} $ so your equation is perfectly valid (if you omit to say you are canceling anything). – N74 Sep 28 '16 at 21:31
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    What if $f$ and $g$ are not transformations of each other, say, $g = x^2/2$ and $f = x + y$? – clueless Sep 28 '16 at 21:36
  • Then you have no chain rule and the equation is not valid. Anyway, in your example, you can still say $g=(f-y)^2/2$ and treat $y$ as a constant. – N74 Sep 28 '16 at 21:39
  • These answers are so snarky, I'm sorry OP. I'd like to know the answer as well, since cancelling derivatives is NOT always allowed. (For example, in @user115350 's answer, if you cancel back the dx and dy's, you get dg/df = 2 dg/df.) – user3932000 Feb 16 '18 at 10:15

1 Answers1

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This is not by accident. As N74 said, this is chain rule. So $$\frac{\partial g}{\partial f}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial f}+\frac{\partial g}{\partial y}\frac{\partial y}{\partial f}=\frac{\frac{\partial g}{\partial x}}{\frac{\partial f}{\partial x}}+\frac{\frac{\partial g}{\partial y}}{\frac{\partial f}{\partial y}}$$

user115350
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    Well, your $\partial g/\partial f = 4xy$ is different though since you included $\partial y$. Now I'm confused. – clueless Sep 28 '16 at 21:50