# 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.

• I wrote "Quotient rule" as the subject line and I was told that's too short. Any mental anguish resulting from the current form of the subject line is regrettted, albeit not very much. $\qquad$ Jul 14, 2017 at 0:29
• How 'bout "A Geometric or Conceptual Interpretation of the Quotient Rule"? Having said that, I believe Gil Strang gives a similar triangles picture for the reciprocal rule in his calculus book, and I've used it in classes a few times. Jul 14, 2017 at 0:33
• @TedShifrin : But what I'm wondering about is an interpration of the numerator in the quotient rule. Jul 14, 2017 at 4:00
• @MichaelHardy Since you are the tag-creator, I thought it could be useful to let you know that I have opened a thread on meta about this tag: What is the (calculus-identities) tag intended for? Jul 2, 2018 at 18:03
• Of possible related interest is this 1 March 2007 ap-calculus post archived at Math Forum. Jul 14, 2018 at 11:21

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}

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

# 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.

• Only after posting, I noticed that the second part is essentially the same as the post of eyeballfrog. However I decided to leave it in because it makes the post more complete. The first part however is not covered by other answers. Jul 21, 2019 at 7:43