I know that if $u=u(s,t)$ and $s=s(x,y)$ and $t=t(x,y)$ then the chain rule is $$\begin{align}\color{blue}{\fbox{$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial s}\times \frac{\partial s}{\partial x}+\frac{\partial u}{\partial t}\times \frac{\partial t}{\partial x}$}}\color{#F80}{\tag{A}}\end{align}$$

A short extract from my book tells me that:

If $u=(x^2+2y)^2 + 4$ and $p=x^2 + 2y$ then $u=p^2 + 4$ therefore $$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial p}\times \frac{\partial p}{\partial x}\tag{1}$$ as $u=u(x,y)$ and $p=p(x,y)$

The book mentions no origin of equation $(1)$ and unlike $\color{#F80}{\rm{(A)}}$ is has only one term on the RHS; So I would like to know how it was formed. Is $(1)$ simply equivalent to $\color{#F80}{\rm{(A)}}$ but with the last term missing? Or is there more to it than that?

Many thanks,


  • 6
    $\begingroup$ Tangential comment: one of my mathematical pet peeves is that, when stating the chain rule, the function $\bar u(x,y) = u(s(x,y), t(x,y))$ is often called $u$. So we have two different functions which are given the same name. (This is happening here in equation (A).) $\endgroup$ – littleO Jan 4 '16 at 6:28
  • $\begingroup$ Possible duplicate of math.stackexchange.com/q/1577104/265466. $\endgroup$ – amd Jan 4 '16 at 6:48
  • 2
    $\begingroup$ Are you sure it said $\partial u/\partial p$ rather than $du/dp$? $\endgroup$ – David Jan 4 '16 at 6:53
  • $\begingroup$ @David It said $du/dp$; but it is equivalent to $\partial u/\partial p$ in the one dimensional case, so it doesn't matter. $\endgroup$ – BLAZE Jan 4 '16 at 7:00
  • 7
    $\begingroup$ In an above comment, you say "it doesn't matter" (partial vs. total d). This is a mathematical pet peeve of mine because many students use this argument to fail to understand that $\frac{\mathrm{d} u}{\mathrm{d} p}$ is the derivative of a dependent variable with respect to an independent variable and $\frac{\partial u}{\partial p}$ is the derivative of a recipe with respect to one of its formal parameters ("slots"). For instance, what do $\frac{\partial u(p,q)}{\partial q}$ ($=0$ in your question) and $\frac{\partial u(p,p)}{\partial p}$ mean? $\endgroup$ – Eric Towers Jan 4 '16 at 7:21

To expand a bit on Hagen von Eitzen’s answer and littleO’s comment, there are really two different functions that are both named ‘$u$’. The first is a function of two variables, $u:(x,y)\mapsto (x^2+2y)^2+4$, while the second is a function of only one variable, $u:t\mapsto t^2+4$. Let’s call the former $\bar u$ to keep them straight. We also have $p:(x,y)\mapsto x^2+2y$, so $\bar u=u\circ p$, i.e., $\bar u(x,y)=u(p(x,y))$. By the chain rule, ${\partial\over\partial x}\bar u={\partial\over\partial x}(u\circ p)=\sum{\partial u\over\partial w_i}{\partial w_i\over\partial x}$, the sum taken over all of the parameters $w_i$ of $u$. In this case, $u$ is a function of only one variable, so this sum has only the one term, ${\partial u\over\partial p}{\partial p\over\partial x}$. Because this $u$ is a function of only one variable, you might see this written as ${du\over dp}{\partial p\over\partial x}$ instead.

  • $\begingroup$ Thanks for your answer, I found this formula: $\frac{\partial y}{\partial x_i} = \sum_{\ell = 1}^m \frac{\partial y}{\partial u_\ell} \frac{\partial u_\ell}{\partial x_i}$ on wikipedia; I was wondering if you could confirm this is the same formula you're using in your answer? Hagen von Eitzen has gone unresponsive; I think the formula that Hagen suggests is inapplicable here since in Hagen's formula there is no partial derivative on the LHS. Could you please demonstrate the part of Hagens answer marked in red? Then I will accept your answer. $\endgroup$ – BLAZE Jan 6 '16 at 11:45
  • $\begingroup$ @BLAZE Yes, the formula you found and what I wrote are equivalent. The bottom line is that the chain rule for partial derivatives requires you to differentiate $y$ with respect to every one of its parameters. Note that in Hagen’s answer, $f$ is a function with only one parameter, $t$, so there’s no practical difference between ${df\over dt}$ and ${\partial f\over\partial t}$ for this function. That’s exactly the case for the function I’ve called $u$. $\endgroup$ – amd Jan 7 '16 at 8:52

More generally, if $u(x_1,\ldots,x_n)$ is a partially differentiable function function in $n$ variables and $s_1,\ldots ,s_n$ are differentiable and $f(t)=u(s_1(t),\ldots,s_n(t))$ then $$\frac {df}{dt}=\frac{\partial u}{\partial x_1} \frac {d s_1}{d t}+\ldots +\frac{\partial u}{\partial x_n} \frac {d s_n}{d t}$$ Your $(\mathrm{A})$ is a special case of $n=2$ and $(1)$ is a special case of $n=1$

  • 3
    $\begingroup$ While correct, this is a bit short. A sentence explaining why $n=1$ in (1) would help. (E.g. like @amd's answer) $\endgroup$ – Stig Hemmer Jan 4 '16 at 9:09
  • 1
    $\begingroup$ @Hagen So for $\color{#F80}{\rm{(A)}}$ $n=2\implies u(x_1,x_2)$ and $f(t)=u(s_1(t),s_2(t))\implies \cfrac {df}{dt}=\cfrac{\partial u}{\partial x_1} \cfrac {d s_1}{d t}+\cfrac{\partial u}{\partial x_2} \cfrac {d s_2}{d t}$ but how is this the same as $\color{blue}{\fbox{$\cfrac{\partial u}{\partial x}=\cfrac{\partial u}{\partial s}\times \cfrac{\partial s}{\partial x}+\cfrac{\partial u}{\partial t}\times \cfrac{\partial t}{\partial x}$}}$? As $\color{#F80}{\rm{(A)}}$ contains all partial derivatives. Also as mentioned by Stig Hemmer; Can you please explain how $(1)$ is a special of $n=1$? $\endgroup$ – BLAZE Jan 4 '16 at 12:10

It comes from the divergence operator $\nabla$. Let $f$ be a scalar valued function then $\nabla f \equiv \partial \left\langle \dfrac{\partial f}{\partial x}, \dots \right\rangle$ vectorizes $f$. If you picture $f$ as the height of a hill and its parameters the coordinates of each point of the hill on the Earth, then $\nabla f$ points in the direction of largest change per unit distance on the Earth surface. So in your example, $s,t$ are the Earth surface coordinates. Now take the dot product with the tangent vector of a curve on the Earth's surface parameterized by $x: \langle s(x), t(x) \rangle$. Figure out what that means by looking at what dot product means.

  • 2
    $\begingroup$ As far as I can tell this is an answer to only the title of the post, rather than the question asked in the body. $\endgroup$ – Eric Wofsey Jan 4 '16 at 6:43
  • $\begingroup$ @EricWofsey That's right, I need to know how equation $(1)$ was formulated (presumably from $\color{#F80}{\rm{(A)}}$ in my opinion). $\endgroup$ – BLAZE Jan 4 '16 at 6:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.