Quotient rule/Quotient rule \begin{align}
& f(x)^2 \frac d {dx}\, \frac{g(x)}{f(x)} = f(x)g'(x) - f'(x)g(x). \tag 0 \\[10pt]
& g(x)^2 \frac d {dx}\, \frac{f(x)}{g(x)} = g(x)f'(x) - g'(x)f(x). \tag 0 \\[10pt]
\text{Therefore } & f(x)^2 \frac d {dx}\, \frac{g(x)}{f(x)} + g(x)^2 \frac d {dx}\, \frac{f(x)}{g(x)} = 0. \tag 1
\end{align}
If one had never heard of the quotient rule, one could of course prove $(1)$ by an argument paralleling the sort used to prove the quotient rule, or by first proving the quotient rule and deducing $(1)$ as a corollary (the latter option being of course what I did).
But:


*

*Is there some natural interpretation of the right side of $(0)$, perhaps not relating it to the left side? (Maybe this part is the main point of this question.)

*Is there a quick intuitive argument for $(1)$ -- thus a simpler argument than any that proves the quotient rule or does something paralleling that proof?

*Similarly for the left side of $(0)$?


Perhaps we should note that $(0)$ is weaker than the quotient rule in that it does not imply differentiability of the quotient, but rather it assumes it.
 A: Assuming you know the product rule, $(1)$ can be proved from
\begin{align}
& f(x)^2\frac{d}{dx}\left[\frac{g(x)}{f(x)}\right]+g(x)^2\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = f(x)g(x)\left(\frac{f(x)}{g(x)}\frac{d}{dx}\left[\frac{g(x)}{f(x)}\right]+\frac{g(x)}{f(x)}\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]\right) \\[12pt]
= {} &  f(x)g(x)\frac{d}{dx}\left[\frac{f(x)}{g(x)}\frac{g(x)}{f(x)}\right] = f(x)g(x)\frac{d1}{dx} = 0
\end{align}
A: That expression, $fg'-f'g$ has this interpretation, of sorts. Assume $f$ and $g$ are positive functions, and consider the sign of our expression:
$$fg'-f'g>0\\ \Leftrightarrow fg'>f'g\\ 
\Leftrightarrow \frac{g'}g > \frac{f'}f\\
\Leftrightarrow \frac{d}{dx}\log g > \frac{d}{dx}\log f$$
If $f$ and $g$ are both functions going to infinity, let's say, this tells you something about their relative growth rates.
Not sure if that helps or not, but nobody had mentioned it yet.....
A: A simple interpretation is that $\, fg'-f'g \,$ is the determinant of a 
$\, 2\times 2 \,$ matrix and when we switch $\, f \,$ and $\, g \,$ the determinant changes sign. Next, the difference quotient for $\, f/g \,$ is the determinant of a $\, 2\times 2 \,$ matrix divided a product of two values of the quotient function. In the limit this becomes the first determinant divided by the square of the quotient function.
Given functions $\, f,g \,$  define $\, h(x) \!:=\! f(x)/g(x). \,$ If $\, x_1 := X_1/Y_1, \, x_2 := X_2/Y_2, \,$ the difference quotient is
 $$ \Delta_{x_1,x_2}(h) := \frac{h(x_1) - h(x_2)}{x_1-x_2} =
 \frac{ f(x_1)/g(x_1) - f(x_2)/g(x_2)} {x_1-x_2} =
 \frac{  f(x_1)g(x_2) - f(x_2)g(x_1) }{(x_1-x_2)g(x_1)g(x_2)}. \,$$
In the limit, as $\, x_1 \,$ and $\, x_2 \,$ both approach $\, x \,$ this becomes $\, (f(x)g'(x) - f'(x)g(x))/g(x)^2. \,$
Using homogeneous coordinates allows us to define a more symmetric version of
 $\, h(x). \,$ Define $\, H([X:Y]) := [f(X/Y):g(X/Y)]. \,$
The symmetric homogeneous version of the difference quotient is
 $$ \Delta_{x_1,x_2}(H) := \Delta_{x_1,x_2}(h) \, \frac{g(x_1)g(x_2)}{Y_1\,Y_2} =
 \begin{vmatrix} f(x_1) & f(x_2) \\ g(x_1) & g(x_2) \end{vmatrix}  \Bigg/ \begin{vmatrix} X_1 & X_2 \\ Y_1 & Y_2 \end{vmatrix}. $$
A: Right side of $(0)$
The right side of $(0)$ is the Wronskian of $f$ and $g$, more exactly, the special case of the Wronskian of two functions. It is used to establish linear independence of solutions to differential equations: It vanishes identically for linearly dependent functions (but may also vanish for independent ones). For two functions, it's quite obvious from the quotient rule: If the two functions are linearly dependent, their quotient is constant, and thus its derivative vanishes.
As a side remark, in one-dimensional non-relativistic quantum mechanics, the state of a particle is described by a function $\psi:\mathbb R\to\mathbb C$, the so-called wave function (the argument being the position in space), which has a probabilistic interpretation. The probability current then is, up to a constant factor, the Wronskian of the wave function and its complex conjugate.
Quick argument for $(1)$
Quite obviously,
$$\frac{f(x)}{g(x)}\,\frac{g(x)}{f(x)} = 1$$
and therefore its derivative is $0$. But according to the product rule,
$$0=\frac{\mathrm d}{\mathrm dx}\left(\frac{f(x)}{g(x)}\,\frac{g(x)}{f(x)}\right)
= \left(\frac{\mathrm d}{\mathrm dx}\frac{f(x)}{g(x)}\right)\frac{g(x)}{f(x)}
+ \frac{f(x)}{g(x)}\left(\frac{\mathrm d}{\mathrm dx}\frac{g(x)}{f(x)}\right)$$
Multiply both sides by $f(x)g(x)$, and you've got equation $(1)$.
So ultimately equation $(1)$ is the derivative applied to fact that the two fractions are inverses of each other.
