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I'm just starting partials and don't understand this at all. I'm told to hold $y$ "constant", so I treat $y$ like just some number and take the derivative of $\frac{1}{x}$, which I hope I'm correct in saying is $-\frac{1}{x^2}$, then multiply by $y$, getting $-\frac{y}{x^2}$.

But apparently the correct answer is $\frac{1}{x}$. What am I missing?

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  • $\begingroup$ @ZenLogic: No, that's the integral :) $\endgroup$ – psmears Mar 20 '15 at 10:14
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    $\begingroup$ Taking the partial derivative with respect to $y$ means holding constant all other variables; here that means holding constant $x$, not $y$. $\endgroup$ – Marc van Leeuwen Mar 20 '15 at 11:12
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When you take the derivative of $\frac{y}{x}$ with respect to $y$ you are computing $\frac{\partial }{\partial y} \frac{y}{x} = \frac{1}{x}$ because here you are holding $x$ constant. If you take the derivative of the same expression with respect to $x$ then you compute $\frac{\partial}{\partial x} \frac{y}{x} = - \frac{y}{x^2}$ and this is when you hold $y$ constant.

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    $\begingroup$ Huh, I just had it backward I guess. Thanks. $\endgroup$ – SquarerootSquirrel Mar 20 '15 at 3:49
  • $\begingroup$ @SquarerootSquirrel: I this answer was helpful to you, you might wish to accept it. $\endgroup$ – Marc van Leeuwen Mar 20 '15 at 11:16
  • $\begingroup$ @අරුණ I'm shocked, your answer is better. I even up-voted yours :) $\endgroup$ – Mnifldz Mar 20 '15 at 14:02
  • $\begingroup$ @Mnifldz Actually I also up voted yours ;) $\endgroup$ – ASB Mar 20 '15 at 14:27
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By first principles,

$ \dfrac{\partial}{\partial y}\left( \dfrac{y}{x}\right)=\lim\limits_{\delta y\to 0}\left( \dfrac{\dfrac{y+\delta y}{x}-\dfrac{y}{x}}{\delta y}\right)=\lim\limits_{\delta y\to 0}\dfrac{\delta y}{x\delta y}=\lim\limits_{\delta y\to 0}\dfrac{1}{x}=\dfrac{1}{x} $

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    $\begingroup$ This answer is so nice as it shows every "fancy" statement comes from first principles. $\endgroup$ – Mahdi Mar 20 '15 at 5:22
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    $\begingroup$ I don't think the confusion of OP came from the definition of the derivative of a function of one variable, so I think this answer just muddles things unnecessarily. $\endgroup$ – Marc van Leeuwen Mar 20 '15 at 11:15
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"With respect to $y$" means that you will be holding $x$ constant, not $y$. $y$ is the variable we are differentiating with respect to, so it is /not/ to be treated as a constant!

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There is nothing special about the symbol $x$. Unfortunately, there's a certain bias to calling $f$ the function and $x$ the variable, but really it should never1 matter how variables are labelled as long as it's done consistently.

So in particular, if you write $$ f(x,y) = \tfrac{y}{x} $$ it means exactly the same thing as $$ f(y,x) = \tfrac{x}{y}. $$ Note that $x$ and $y$ aren't actually part of the definition: this equation defines only $f$, and introduces two new “private” symbols, locally, to do that. I could also have written $$ f(\mathscr{Y},\Xi) = \tfrac{\Xi}{\mathscr{Y}}. $$

Now, what you're doing is actually $$ f'(x,y) \equiv \tfrac{\partial}{\partial y} f(x,y), $$ and that again is the same as $$ f'(y,x) \equiv \partial_x f(y,x) = \partial_x \tfrac{x}{y}. $$ You certainly won't doubt that this is $\tfrac1y$... although again that statement is meaningless without context: really I should say that $f'(y,x) = \tfrac1y$, and therefore $f'(x,y) = \tfrac1x$.


1Alas, in many applications of maths this is widely neglected! In physics, two equations may be interpreted as something completely different depending on how the variables are labelled.

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You're supposed to hold x constant.

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have you looked at the graph? if you have a good enough plotting program (good enough to rotate 3D graphs), it should be easy to see that the surface is linear for all fixed x (and hyperbolic for all fixed y)

first pass:

http://www.wolframalpha.com/input/?i=plot+y%2Fx+for+x%3D-1..1%2C+y%3D-1..1

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