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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?

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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. Aug 8 '12 at 8:40
You might profit from this answer. – joriki Aug 8 '12 at 8:42
@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 8 '12 at 9:00
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. Aug 8 '12 at 9:02

1 Answer 1

up vote 1 down vote accepted

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

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

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What is variable $z$? – Michael Litvin Aug 8 '12 at 8:56
@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 8 '12 at 9:00
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 8 '12 at 9:11
@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 8 '12 at 9:21
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 at 0:38

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