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.
Best Answer
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 $n=1$