This question is a simplified version of this previous question asked by myself.

The following is a short extract from a book I am reading:

If $u=(x^2+2y)^2 + 4$ and $p=x^2 + 2y$ $\space$ then $u=p^2 + 4=f(p)$ therefore $$\frac{\partial u}{\partial x}=\frac{\rm d f(p)}{\rm d p}\times \frac{\partial p}{\partial x}=2xf^{\prime}(p)\tag{1}$$ and $$\frac{\partial u}{\partial y}=\frac{\rm d f(p)}{\rm d p}\times \frac{\partial p}{\partial y}=2f^{\prime}(p)\tag{2}$$

I know that the chain rule for a function of one variable $y=f(x)$ is $$\begin{align}\color{red}{\fbox{$\frac{{\rm d}}{{\rm d}x}=\frac{{\rm d}}{{\rm d}y}\times \frac{{\rm d}y}{{\rm d}x}$}}\color{red}{\tag{A}}\end{align}$$

I also know that if $u=f(x,y)$ then the differential is

$$\begin{align}\color{blue}{\fbox{${{\rm d}u=\frac{\partial u}{\partial x}\cdot{\rm d} x+\frac{\partial u}{\partial y}\cdot{\rm d}y}$}}\color{blue}{\tag{B}}\end{align}$$

I'm aware that if $u=u(x,y)$ and $x=x(t)$ and $y=y(t)$ then the chain rule is $$\begin{align}\color{#180}{\fbox{$\frac{\rm d u}{\rm d t}=\frac{\partial u}{\partial x}\times \frac{\rm d x}{\rm d t}+\frac{\partial u}{\partial y}\times \frac{\rm d y}{\rm d t}$}}\color{#180}{\tag{C}}\end{align}$$

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

Could someone please explain the origin or meaning of equations $(1)$ and $(2)$?

The reason I ask is because I am only familiar with equations $\color{red}{\rm (A)}$, $\color{blue}{\rm (B)}$, $\color{#180}{\rm (C)}$ and $\color{#F80}{\rm (D)}$ so I am not used to seeing partial derivatives mixed up with ordinary ones in they way they were in $(1)$ and $(2)$.

Many thanks,


  • 2
    $\begingroup$ @BLAZE Would it make more sense to you if it were written as $\frac{\partial u}{\partial x}=\frac{\partial f}{\partial p}\times \frac{\partial p}{\partial x}$? Because partial derivatives and ordinary derivatives are equivalent in the one-dimensional case. $\endgroup$
    – user137731
    Dec 30, 2015 at 20:42

2 Answers 2


This is often confusing because there is a conflation of the symbol for a function argument as opposed to a function itself. For example, when you write $f(p) = p^2 + 4$, you are thinking of $f$ as a function and $p$ as the argument of that function, which could be any dummy variable. In fact, let us write $f(\xi) = \xi^2+4$, which is the same function with simply another symbol representing the rule that $f$ implies. At the same time, you are using the symbol $p$ as a function $p(x,y) = x^2 + 2y$. Now, with the functions $f(\xi)$ and $p(x,y)$ you have $u(x,y) = f \circ p\,(x,y)$; that is, $u$ is the composition of $f$ with $p$. Hence, using the chain rule and suppressing the variables $x$ and $y$, you have $$ \frac{\partial u}{\partial x} = f'(p)\, \frac{\partial p}{\partial x} = \frac{d f}{d \xi} (p) \, \frac{\partial p}{\partial x} $$ where $f' = \frac{d f}{d \xi}$ since we changed the argument symbol of $f$ to $\xi$ -- notice that $\frac{d f}{d\xi}$ is still evaluated at the function $p$ by the chain rule. If we wanted to to explicitly show where the variables $x$ and $y$ would manifest, we would have $$ \frac{\partial u}{\partial x}(x,y) = f'(p(x,y))\, \frac{\partial p}{\partial x}(x,y) = \frac{d f}{d \xi} (p(x,y)) \, \frac{\partial p}{\partial x}(x,y). $$ Hopefully this helps.


The given function $u$ has two variables $x$ and $y$. So, it makes sense to talk about partial derivatives $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$. While you take another 'variable' $p$ for $u$, then you have a function of one variable, but which is dependent of other two variables, and so is $u$ then.

In other words you have: $$\frac{\partial u(p(x,y))}{\partial x}=\frac{\partial u(p)}{\partial p}\frac{\partial p(x,y)}{\partial x}$$ and $$\frac{\partial u(p(x,y))}{\partial y}=\frac{\partial u(p)}{\partial p}\frac{\partial p(x,y)}{\partial y}.$$

  • 1
    $\begingroup$ Thanks for your answer. When you write $\cfrac{\delta u}{\delta x}$ do you mean $\cfrac{\partial u}{\partial x}$ as $\delta$ is not the same as $\partial$? $\endgroup$
    – BLAZE
    Dec 30, 2015 at 20:41
  • $\begingroup$ Much better thanks. $\endgroup$
    – BLAZE
    Dec 30, 2015 at 20:47

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