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.

  • $\begingroup$ By the way, in your equation for $\frac{dF}{dt}$, two different functions are both called the same name $F$. The $F$ on the left is not the same function as the $F$ on the right. This is a common abuse of notation, but I think it sometimes causes confusion. $\endgroup$
    – littleO
    Commented Jul 21, 2018 at 19:03
  • $\begingroup$ @littleO Thanks for note, I'll edit it. $\endgroup$ Commented Jul 21, 2018 at 19:05
  • $\begingroup$ F(x,y)= x+ y cannot be written as "$F_1(x)F_2(y)$". $\endgroup$
    – user247327
    Commented Jul 21, 2018 at 19:13
  • $\begingroup$ It is really not clear what you are asking here. How can $F$ have a variable number of parameters? $\endgroup$
    – copper.hat
    Commented Jul 21, 2018 at 19:23

1 Answer 1


There are functions that are not separable. For example: $F(x,y)=\sin(x y)$

  • $\begingroup$ But Why it is possible to derive the chain rule formula if $F(x,y)$ can be written as product?? $\endgroup$ Commented Jul 21, 2018 at 19:17
  • $\begingroup$ When you say: "we can plug F1 in the derivative without affecting it doing that will give us :" you obtain: $$F_1 \frac{d F_2}{dx} = \frac{\partial{\left(F_1 F_2 \right)}}{\partial{x}} = \frac{\partial{F}}{\partial{x}}$$ and this works $\endgroup$ Commented Jul 21, 2018 at 19:32
  • $\begingroup$ You have not made it clear why functions that are not separable answers the question. Also how did $F_1 \frac{d F_2}{dx} $ become $\frac{\partial{\left(F_1 F_2 \right)}}{\partial{x}}$, in the first instance $F_1$ is not being differentiated wrt to $x$, and then it is? $\endgroup$ Commented Jul 21, 2018 at 20:46
  • $\begingroup$ $F_1$ behaves like a constant $\endgroup$ Commented Jul 22, 2018 at 7:07

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