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,
BLAZE.