So what I'm trying to do here is proving the chain rule of differentiation several variable function , but I'll work here with a simpler case a function of to variable , I want to check my proof.
First let $$F(x,y) $$ be a function of $x$ and $y$ such that $$F(x,y) = F(x(t),F(y(t))$$
then its derivate with respect to $t$ is the following:
$$\frac{dF}{dt} = \frac{\partial F}{\partial x} \frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}$$
now consider the following let's break $F(x,y) $ into
$$F(x,y)=F_{1}(x)F_{2}(y)=F_{1}(x(t))F_{2}(y(t))$$
now differentation by product rule gives us
$$F_{1}\frac{dF_{2}}{dt}+F_{2}\frac{dF_{1}}{dt}$$
using ordinary single variable chain rule for both $F_{2}$ and $F_{1}$ yield the following :
$$F_{1}\frac{dF_{2}}{dx}\frac{dx}{dt}+F_{2}\frac{dF_{1}}{dy}\frac{dy}{dt}$$
now the next step is to observe that $F_{1}$ is a constant with respect to $F_{2}$ and vice versa is true, That mean we can plug $F_{1}$ in the derivative without affecting it doing that will give us : $$F_{1}\frac{dF_{2}}{dx}\frac{dx}{dt}+F_{2}\frac{dF_{1}}{dy}\frac{dy}{dt} = \frac{dF}{dt}$$ which is the chain rule for two variable function $F(x,y)$ , what I want to do here is to prove that every two variable $F(x,y)$ , or more , can be written as product of single variable , or simply I want to prove that $F(x,y)$=$F_{1}(x)F_{2}(y)$ for every $F(x,y)$ ,
last note $F(x,y)$ is differentiable everywhere.
Best Answer
There are functions that are not separable. For example: $F(x,y)=\sin(x y)$