Product rule for independent functions Suppose we have two independent functions $f(u)$ and $g(v)$ then
(i) :- $$\frac {d \;(f(u)\cdot g(v))}{dx} = \frac {d (f(u)\cdot g(v))}{du}\cdot\frac{du}{dx} $$
Since $g(v)$ doesn't change with changing $u$ $$\frac {d (g(v))}{du} = 0$$
So the first equation becomes
(ii) :- $$\frac {d \;(f(u)\cdot g(v))}{dx} = g(v) \cdot f'(u) \cdot\frac{du}{dx} $$
Similarly we can get this equation
(iii) :- $$\frac {d \;(f(u)\cdot g(v))}{dx} = f(u)\cdot g'(v)\cdot \frac{dv}{dx} $$
Now if I add both the equations I will get
$$ \frac {d \;(f(u)\cdot g(v))}{dx} = \left( \frac{1}{2} \right) \left(g(v)\cdot f'(u)\cdot\frac{du}{dx} + f(u)\cdot g'(v)\cdot\frac{dv}{dx}\right)$$
But this can't be right since it is half of what we get from the product rule.
So where am I wrong ?
 A: Your derivation is based on the assumption that, $$\dfrac{d\left(f(u)g(v)\right)}{dx}=\dfrac{d\left(f(u)g(v)\right)}{du}\dfrac{du}{dx}=\dfrac{d\left(f(u)g(v)\right)}{dv}\dfrac{dv}{dx}$$
But,
$$
\dfrac{d\left(f(u)g(v)\right)}{du}\dfrac{du}{dx}\;\text{ doesn't necessarily be equal to }\;\dfrac{d\left(f(u)g(v)\right)}{dv}\dfrac{dv}{dx}.$$
For example, let $f(u) = u^2, g(v) = v^3+v, u(x) = \sin (x), v(x) = x+1$. 
Here,
$$
\dfrac{d\left(f(u)g(v)\right)}{du}\dfrac{du}{dx} =g(v)\dfrac{d\left(f(u)\right)}{du}\dfrac{du}{dx}= (v^3+v)(2u)\cos(x)
$$
but,
$$
\dfrac{d\left(f(u)g(v)\right)}{dv}\dfrac{dv}{dx} =f(u)\dfrac{d\left(g(v)\right)}{du}\dfrac{dv}{dx}=u^2(3v^2+1)(1) 
$$
And both aren't the same.

What you have to do is,
$$
\dfrac{d\left(f(u)g(v)\right)}{dx} = g(v)\dfrac{d\left(f(u)\right)}{dx} + f(u)\dfrac{d\left(g(v)\right)}{dx}=g(v)\dfrac{d\left(f(u)\right)}{du}\dfrac{du}{dx} + f(u)\dfrac{d\left(g(v)\right)}{dv}\dfrac{dv}{dx}$$

Moreover, if
$$
\begin{split}& g(v)\dfrac{d\left(f(u)\right)}{du}\dfrac{du}{dx}=f(u)\dfrac{d\left(g(v)\right)}{dv}\dfrac{dv}{dx}\\
\implies&\dfrac{f'(u)du}{f(u)} =\dfrac{g'(v)dv}{g(v)}\\
\implies& \ln|f(u)| = \ln|kg(v)| \\
\implies& \boxed{f(u) = kg(v)} 
\end{split}
$$

But $f(u)$ and $g(v)$ are dependent only on $u$ and $v$ respectively.


Both $f(u)$ and $g(v)$ have to be constant functions for this condition to occur and in this case the total derivative will be zero, as $f'(u)$ and $g'(v)$ are zeroes.
So in this particular condition, it doesn't matter whether you have half the formula or the exact formula.

A: Let us make the implicit assumptions explicit, you are assuming that $u(x)$ and $v(x)$ which means that as you change a parameter $x$, there are two other functions which change according to it.
Physical example: You pump in air (volume) into a balloon, then as you increase it, both the mass of the balloon and density of balloon changes. Now, you may have other two function dependent on mass and density of balloon.
Now, we need to compute $\frac{d}{dx} \left[ g(v) f(u) \right]$ , by the multivariable chain rule:
$$ \frac{d}{dx} g(v) f(u) = \frac{\partial g}{\partial v}  \frac{dv}{dx} f(u) + \frac{\partial f}{\partial u} \frac{du}{dx} g(v)$$
To reduce this to the single variable product rule we consider the fact that $g$ is only a function of $v$ and $f$ is only a function of $u$:
$$ \frac{d}{dx} g(v) f(u) = g'(v) f(u) \frac{dv}{dx} + f'(u) \frac{du}{dx} g(v)$$
The general MV chain rule for two variable is given as:
$$ \frac{d}{dt} F(u(t) , v(t) ) = \frac{\partial F}{\partial u} \frac{du}{dt} + \frac{\partial F}{\partial v} \frac{dv}{dt}$$
A: I think that this statement is wrong:

Since $g(v)$ doesn't change with changing $u$

if $u=\phi(x)$, then we can probably partition its domain in such a way that we can always find $x=\phi^{-1}(u)$ for a given partition.
since $v=\psi(x)$ and $x=\phi^{-1}(u)$, we get $v=\psi(\phi^{-1}(u))=\xi(u)$.
A: If your method were right, you could also subtract your partial result to arrive at
$$g(v) \cdot f'(u) \cdot\frac{du}{dx}-f(u)\cdot g'(v)\cdot \frac{dv}{dx}=0$$
which is certainly not generally true.
What your partial results actually compute, is in fact not $\frac {d \;(f(u)\cdot g(v))}{dx}$ in both cases, but only  a partial contribution to$\frac {d \;(f(u)\cdot g(v))}{dx}$, and you have to really add these contributions, not take their average.

By the end of the day, this (or the product rule per se) is just the (multidimensional) chain rule, which circumvents the magic of keeping partial results and combining them by having several dimensions available; in short, the derivative of $\pmatrix{x_1\\\vdots\\x_n}\mapsto \pmatrix{f_1(x_1,\ldots,x_n)\\\vdots\\f_m(x_1,\ldots,x_n)}$ is $\pmatrix{\frac{\partial}{\partial x_1}f_1&\cdots&\frac{\partial}{\partial x_n}f_1\\\vdots&\ddots&\vdots\\\frac{\partial}{\partial x_1}f_m&\cdots&\frac{\partial}{\partial x_n}f_m}$ and we conveniently identify $1\times 1$ matrices with scalars:
We have the map sequence
$$ x\mapsto \pmatrix{u(x)\\v(x)}\mapsto \pmatrix{f(u(x))\\g(v(x))}\mapsto f(u(x))g(v(x)).$$

*

*The derivative of the first step $x\mapsto\pmatrix{u(x)\\v(x)}$ is $\pmatrix{u'(x)\\v'(x)}$,

*the derivative of the second step $\pmatrix{u\\v}\mapsto \pmatrix{f(u)\\g(v)}$ is $\pmatrix{f'(u)&0\\0&g'(v)}$ (where the zeroes indicate the "independent functions" property), or evaluated at $\pmatrix{u(x)\\v(x)}$: $\pmatrix{f'(u(x))&0\\0&g'(v(x))}$

*the derivative of the last step $\pmatrix{p\\q}\to pq$ is $(q\;p)$, or evaluated at $\pmatrix{f(u(x))\\g(v(x))}$: $(g(v(x))\;f(u(x)))$.

Multiplying all out gives
$$ (g(v(x))\;f(u(x)))\pmatrix{f'(u(x))&0\\0&g'(v(x))}\pmatrix{u'(x)\\v'(x)}=
g(v(x))f'(u(x))u'(x)+f(u(x))g'(v(x))v'(x)$$
