Say I have a function of $n$ variables $F(x_{1}, x_{2}, x_{3},...,x_{n})$, where $x_{1} = g_{1}(y_{1}, y_{2}, y_{3},...,y_{m})$, $x_{2} = g_{2}(y_{1}, y_{2}, y_{3},...,y_{m}),\dots, x_{n} = g_{n}(y_{1}, y_{2}, y_{3},...,y_{m})$ are also some functions of $m$ variables. How can I find the expression for

$$\frac{\partial F}{\partial x_k}$$

in terms of the variables $y_i$. I found in another answer a formula for computing partial derivatives of $F$ with respect to $y_{i}, i=1,...,n $ using the formula:

$$\frac{\partial F}{\partial y_{i}} = \frac{\partial F}{\partial x_{1}}\frac{\partial x_{1}}{\partial y_{i}} + \frac{\partial F}{\partial x_{2}}\frac{\partial x_{2}}{\partial y_{i}} + \ldots + \frac{\partial F}{\partial x_{n}}\frac{\partial x_{n}}{\partial y_{i}} = \sum_{k=1}^{n} \frac{\partial F}{\partial x_{k}}\frac{\partial x_{k}}{\partial y_{i}}$$

which I think is just an application of the total derivative followed by the chain rule, ¿am I right? However, that's not what I'm looking for as you can see.

  • $\begingroup$ You are mixing up things. Just compute $\frac {\partial F} {\partial _{x_k}}$ and evaluate the result in $g_1 (y_1, \cdots, y_m), \cdots, g_n(y_1, \cdots, y_n)$ and your result will be expressed in terms of $y_1, \cdots, y_n$. $\endgroup$
    – Alex M.
    May 24, 2015 at 20:13
  • $\begingroup$ @AlexM. Thanks for your comment, the problem is I can't compute that, there's no specific definition for $F$, so I was looking on some kind of 'chain rule', to expand that in terms of the $y_i$. Check my comment in Chappers answer to see a particular example of this problem. $\endgroup$
    – F.Webber
    May 24, 2015 at 20:34

1 Answer 1


$\partial F/\partial x_k$ just means $$\lim_{h \to 0} \frac{F(x_1,\dotsc,x_{k-1},x_k+h,x_{k+1},\dotsc)-F(x_1,\dotsc,x_{k-1},x_k,x_{k+1},\dotsc)}{h},$$ if it exists, so $g$ and $y$ don't actually enter into it. It's not clear how to talk about this partial derivative in any other way.

(This is one of the problems with partial derivative notation if it's not clear what depends on what and have multiple "layers" of functions. You can write $ D_k F $ for "the derivative of $F$ with respect to its $k$th argument" to avoid this.)

  • $\begingroup$ Thanks for the answer, and excuse me for the delay. I also understood $D_kF$ as the derivative of $F$ with respect to its $k$th argument, but then, how would you compute that expression if a certain argument is a function of other variables, e.g., $\frac{\partial F(x,y)}{\partial x}$ with $x=r\cos\theta$ and $y=r\sin\theta$? It would, according to that definition, be something along the lines of $\frac{\partial F(r\cos\theta,r\sin\theta)}{\partial(r\cos\theta)}$, but that doesn't make any sense to me! Also, there's no definition for $F$ so I can't compute on beforehand and then evaluate. $\endgroup$
    – F.Webber
    May 24, 2015 at 20:32
  • 1
    $\begingroup$ Ah, well in that case, you'll probably have to live with using the chain rule in the form $ \frac{\partial F}{\partial x} = \frac{\partial F}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial F}{\partial \theta} \frac{\partial \theta}{\partial x} $. $\endgroup$
    – Chappers
    May 24, 2015 at 21:05
  • $\begingroup$ Oh, I hadn't thought on applying the chain rule that way, thanks for the answer, problem solved :) $\endgroup$
    – F.Webber
    May 24, 2015 at 21:10

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