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

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.
