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Suppose $f=f(x,y(x))$.

Then applying the chain rule we get $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}$.

From this it seems that it always holds that $\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}=0$.

Where's the mistake?

J. M. ain't a mathematician
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Michael Litvin
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    You seem to be misapplying it, and your confusion is exacerbated by your choice of variables. If you have some two-argument function $f(u,v)$, then $$\frac{\partial}{\partial x}f(x,y(x))=f_u(x,y(x))+f_v(x,y(x))y^\prime(x)$$ where $f_u$ and $f_v$ are the appropriate partial derivatives. – J. M. ain't a mathematician Aug 08 '12 at 08:40
  • You might profit from this answer. – joriki Aug 08 '12 at 08:42
  • @J.M. If I understand correctly, you're basically saying that on the left side I used partial derivative instead of total? (meaning that it should have been ($\frac{df}{dx}$ rather than $\frac{\partial f}{\partial x}$) – Michael Litvin Aug 08 '12 at 09:00
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    It doesn't matter here, since you only have one variable. Your error is in applying the chain rule to the outermost function of two variables. – J. M. ain't a mathematician Aug 08 '12 at 09:02

1 Answers1

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As usual when there's confusion about partial derivatives, everything is readily cleared up if we remedy the deficiency in our notation for them by marking which variables are being held fixed:

$$ \def\part#1#2#3{\left.\frac{\partial #1}{\partial #2}\right|_{#3}} \part fxz=\part fxy\part xxz+\part fyx\part yxz=\part fxz=\part fxy+\part fyx\part yxz\;, $$

so there's no such implication, since

$$ \part fxz\ne\part fxy\;, $$

unless of course you choose $z=y$, in which case indeed

$$ \part yxz=\part yxy=0\;. $$

joriki
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  • What is variable $z$? – Michael Litvin Aug 08 '12 at 08:56
  • @Michael: You tell me! :-) When you write $\partial f/\partial x$, that implies that there's some variable other than $x$ that you're keeping constant. You got around making it explicit by using incomplete notation, but I had to give it a name, so I called it $z$. But I didn't introduce it, it's implicit in what you wrote. – joriki Aug 08 '12 at 09:00
  • I think I got the answer to the original question, but still can't tell what is $z$.. For example $f(x,y)=x+y=x+x^2$, where's $z$? – Michael Litvin Aug 08 '12 at 09:11
  • @Michael: Again, if you don't know what $z$ is, then it made no sense to write $\partial f/\partial x$ in the first place. You must have had something in mind that you wanted to keep fixed while changing $x$; $z$ is just my name for whatever you intended to keep fixed. If you weren't intending to keep any particular quantity fixed, then you had no business writing $\partial f/\partial x$. – joriki Aug 08 '12 at 09:18
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    @Michael: Since you insist on not knowing $z$, I'm wondering whether you did actually mean $\mathrm df/\mathrm dx$. In that case, note that the correct relation would then be

    $$ \def\pa{\mathrm d} \frac{\pa f}{\pa x}=\frac{\partial f}{\partial x}\frac{\pa x}{\pa x}+\frac{\partial f}{\partial y}\frac{\pa y}{\pa x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\pa y}{\pa x};, $$

    with the total derivatives cascading just like the partial derivatives with fixed $z$ did in the other case.

     – joriki Aug 08 '12 at 09:21
  • J.M. is right that you could use the partial derivative instead of the total derivative for $f$ considered as a function of one variable $x$, but mixing that with partial derivatives of $f$ considered as a function of two variables, and then not even marking which variable is being held fixed, would be inviting utter chaos. – joriki Aug 08 '12 at 09:28
  • If you really wanted to do that, you could write

    $$ \def\pa{\partial} \frac{\pa f}{\pa x}=\left.\frac{\partial f}{\partial x}\right|_y\frac{\pa x}{\pa x}+\left.\frac{\partial f}{\partial y}\right|_x\frac{\pa y}{\pa x}=\left.\frac{\partial f}{\partial x}\right|_y+\left.\frac{\partial f}{\partial y}\right|_x\frac{\pa y}{\pa x};, $$

    with partial derivatives of the two-variable $f$ indicating which variable is being held fixed and partial derivatives of the one-variable $f$, a.k.a. total derivatives, not. Still that seems unnecessarily confusing; I'd stick with the total derivative notation.

     – joriki Aug 08 '12 at 09:31
  • Total derivatives notation is exactly what I needed for my problem! thanks :) – Michael Litvin Aug 08 '12 at 10:28
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    Finally, after 3 months, someone have an answer to what is basically the same question I have asked 3 months ago. I should favourite it and check my calculations now – Secret May 25 '15 at 00:38