Why is there only one term on the RHS of this chain rule with partial derivatives?

Question

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,

BLAZE.

Answer 1

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.

Answer 2

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 $n=1$

Answer 3

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.