From high school maths I know the chain rule as$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}.$$

So if I wanted to differentiate $y=\cos x^{2}$ I would set $u=x^{2}$ and $y=\cos u$.

In this Physics Stack Exchange answer (https://physics.stackexchange.com/questions/120007/why-do-we-need-a-metric-to-define-gradient) Christoph's answer states:$$\frac{\partial f}{\partial x^{i}}=\sum_{j}\frac{\partial f'}{\partial x'^{j}}\frac{\partial\phi^{j}}{\partial x^{i}},$$where $x'=\phi(x)$ and $f=f'\circ\phi$. Is this another version of the chain rule, and if so how does it relate to my version?

In the versions of the multivariable chain rule I've seen there are similar/identical terms in the numerator and denominator. Like this: $$\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u}.$$


  • 1
    $\begingroup$ Yes, it's the multivariable version of the chain rule. $\endgroup$
    – md2perpe
    Commented Apr 16, 2019 at 17:13
  • $\begingroup$ Please see my "multivariable" edit. $\endgroup$
    – Peter4075
    Commented Apr 16, 2019 at 17:35
  • $\begingroup$ Well, yes: as you wrote, $x'=\phi(x)$, so they really are equal as in the formula you have at the end. $\endgroup$
    – Brightsun
    Commented Apr 16, 2019 at 17:54

1 Answer 1


When we write $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx},$ in the first factor on the right hand side we treat $u$ as a variable, but in the second factor we treat $u$ as a function. This is a bit troublesome mathematically. For this reason it would be somewhat better to introduce a two functions $f(u)$ and $g(x)$ such that $y(x) = f(g(x))$ and say $$\frac{dy}{dx} = \frac{df}{du} \frac{dg}{dx}.$$

This is what Christoph has done in a multivariable setting.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .