Quotient rule/Quotient rule

Question

\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.

Answer 1

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}

Answer 2

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.....

Answer 3

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 $$ \Delta_{x_1,x_2}(H) := \Delta_{x_1,x_2}(h) \, \frac{g(x_1)g(x_2)}{Y_1\,Y_2} = \begin{vmatrix} f(x_1) & f(x_2) \\ g(x_1) & g(x_2) \end{vmatrix} \Bigg/ \begin{vmatrix} X_1 & X_2 \\ Y_1 & Y_2 \end{vmatrix}. $$

Answer 4

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.