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Context: first year didactics of mathematics course for middle school teacher students (in Norway).

I have a reasonable visualization for the product rule of derivatives: Consider a rectangle with sides $a$ and $b$, change them by $\Delta a$ and $\Delta b$ respectively, take the difference of the areas between the original and the altered rectangle. If desired, next consider parametrized side lengths $f(x)$ and $g(x)$ and so on.

To me, this shows nicely why the cross terms are there in the product rule. It can lead to further discussions of sensitivity of product-type quantities to changes in the factors; changes in the smaller one matter more.

Does a helpful visualization exist for the quotient rule?

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    $\begingroup$ Related: See the sci.math thread Long division initiated by Quentin Grady (28 Feb - 2 Mar 2007). In particular, my 2 March 2007 post discusses three algebraic methods that I titled as: (1) METHOD 1: WORKING DIRECTLY FROM (delta y/x) - y/x (2) METHOD 2: RATIONALIZING THE DENOMINATOR OF (delta y/x) [3] METHOD 3: LONG DIVISION APPLIED TO (delta y/x) The 3rd method is what Quentin Grady's initial post asked about. $\endgroup$ Commented Apr 17 at 18:27
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    $\begingroup$ I always thought the quotient rule was just a special case of the product rule and chain rule for computational ease. It has never struck me as anything important theoretically. $\endgroup$
    – qwr
    Commented Apr 19 at 15:02
  • $\begingroup$ $$ \begin{align} w & = \frac uv \\ {} \\ wv & = u \\ {} \\ w'v+v'w & = u' \\ {} \\ \end{align} $$ Then via routine algebra: $$ w' = \frac{u'v - v'u}{v^2}. $$ The one thing that this does not give you, that the proof via limits does give you, is that conclusion that $w$ is differentiable. But this doesn't seem like a "visualization". $\endgroup$ Commented Jun 12 at 21:28
  • $\begingroup$ @qwr Note my comment above, about that this algebraic argument does not give you. $\endgroup$ Commented Jun 12 at 21:29
  • $\begingroup$ @MichaelHardy when I learned calculus in middle and high school, we barely proved anything. It was just computational and I'm not sure what the use of learning that was. I would've preferred public schools teach statistics and basic economics. $\endgroup$
    – qwr
    Commented Jun 13 at 3:50

7 Answers 7

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Depending on how much algebra you allow, you could make the exact same rectangle picture but label the sides $g(x)$ and $q(x)$ with area $f(x)$. This geometrically enforces $g(x)q(x) = f(x)$, aka $q(x) = \frac{f(x)}{g(x)}$.

This gives the approximation

$$ \Delta f \approx q(x) \Delta g + g(x) \Delta q $$

So

$$ \begin{align*} \Delta q &\approx \frac{\Delta f - q(x)\Delta g}{g(x)}\\ &= \frac{\Delta f - \frac{f(x)}{g(x)}\Delta g}{g(x)}\\ &= \frac{g(x)\Delta f - f(x)\Delta g}{(g(x))^2}\\ \end{align*} $$

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    $\begingroup$ This is a clever and nice answer. It brings to attention that the quotient rule is just an algebraic reframing of the product rule. $\endgroup$
    – user52817
    Commented Apr 16 at 15:22
  • $\begingroup$ @user52817 : It's not quite an algebraic reframing of the product rule, because that reframing only tells you that if $u/v$ is differentiable, then its derivative is a certain thing. The reframing does not tell you that $u/v$ is differentiable. The proof via limits does tell you that. $\endgroup$ Commented Jun 13 at 18:11
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Here are two geometric ways of thinking about the quotient rule. The first is essentially a geometric interpretation of an algebraic manipulation of the product rule. The second is an interpretation of the quotient rule as it is usually written.

Consider a rectangle with length $x$ and height $y$, with area $A$. We want to determine the change in height $\Delta y$ in response to a change in length $\Delta x$ and/or a change in area $\Delta A$.

If we hold $x$ constant and increase $A$ by $\Delta A$, then the resulting change in $y$ is $\Delta y = \frac{\Delta A}{x}$.

A drawing of a rectangle with length x, height y, and area A. Another rectangle drawn above the original rectangle represents an increase in area delta-A and an increase in height delta-y.

If we hold $A$ constant and increase $x$ by $\Delta x$, then the resulting decrease in $y$ is $\Delta y \approx - \frac{y\Delta x}{x}$. It's actually $\Delta y = -\frac{(y-\Delta y)\Delta x}{x}$, but we neglect the $\Delta x \Delta y$ term as usual. A drawing of a rectangle with length x, height y, and area A. Another rectangle drawn to the right of the original rectangle shows an increase in area delta-A and an increase in length delta-x. A rectangle drawn on the original rectangle shows a decrease in area minus delta-A and a decrease in height delta-y.

When $A$ and $x$ change simultaneously, a reasonable approximation for the change in $y$ is $$\begin{align*}\Delta y &\approx \frac{\Delta A}{x} - \frac{y\Delta x}{x}\\&=\frac{\Delta A}{x} - \frac{A\Delta x}{x^2}\\&=\frac{x\Delta A - A\Delta x}{x^2}.\end{align*}$$

We can interpret this last line geometrically if we extrude our rectangle into a square prism with side lengths $x$ and height $y$. The area of each vertical face is $A=xy$. Let's consider the changes in volume that result from changing $x$, $y$, and $A$ on the front face.

A drawing of a rectangular prism with a square base with side lengths x and height y. The resulting increase in volume due to an increase in length delta-x and increase in height delta-y are depicted.

If we fix the height and vary the length of the front face by $\Delta x$, the volume of the prism will increase by $A\Delta x$. If we fix the length of the front face and vary the height by $\Delta y$, the volume will increase by $x^2\Delta y$. Changes in $x$ and $y$ will result in some change in the area of the front face. If we are given that the area of the front face changes by $\Delta A$, then the change in the volume of the prism is $x\Delta A$. We can think of the numerator of the quotient rule as representing the relationship between these changes in volume.

A drawing of prisms showing that x squared times delta-y is approximately x times delta-A minus A times delta-x.

$$x^2\Delta y \approx x \Delta A - A \Delta x$$

(It's for convenience that we only vary the length of the front face and hold the length of the side face constant. If both are allowed to vary, then we will get the same result after accounting for an additional volume change of $A\Delta x$ in the third dimension.)

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  • $\begingroup$ Could you please explain little bit more about x ΔA , if you keep x fixed you need to change y to change A by ΔA . $\endgroup$ Commented Apr 17 at 17:56
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    $\begingroup$ @JanakaRodrigo Yes, if $x$ is fixed, then $y$ must change for $A$ to change. For the volume interpretation, I didn't consider $x$ to be fixed while $A$ was varying, unlike for the area interpretation. I've tried to make this clearer. You could also reason about the volumes by thinking about the change in $y$ when $A$ varies and $x$ is fixed and when $x$ varies and $A$ is fixed, as I did with the areas. I wanted to try to show both types of reasoning. $\endgroup$ Commented Apr 18 at 2:23
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slope If you don't mind using similar triangles and are comfortable with both derivatives positive, you can just set $$ OA=g(x), OC=f(x), CD=XZ=\Delta f(x), AB=ZT=\Delta g(x) $$ and write $$ \frac {f+\Delta f}{g+\Delta g}=\frac{BT}{OB}=\frac{AY}{OA} \\ =\frac{AX+XZ-YZ}{OA}=\frac{AX}{OA}+\frac{XZ}{OA}-\frac{YZ}{OA} $$ The first term is $\frac fg$, the second one is $\frac{\Delta f}g$ and the main difficulty is to discern the meaning of the third (subtracted) term. By the similarity of $OAY$ and $YZT$, we have $$ ZY=AY\frac{ZT}{OA}=AY\frac{\Delta g}g $$ and now it boils down to how much hand-waving you are comfortable with to say that $AY$ is essentially $f$ (on the picture it is conveniently between $f$ and $f+\Delta f$ but it won't be so for different choices of signs).

Generally I often prefer a completely different route, however, which goes along with the mantra that for the addition/subtraction one should add/subtract absolute errors but for multiplication/division one should add/subtract relative ones as a first order approximation. That story can be told before introducing the formal notion of the derivative or even of the limit, though, of course, the related computations and pictures are pretty much the same. Once it is firmly in place (you can choose the level of rigor that best suits your needs), the product and quotient rules become simple consequences for arbitrarily long products/quotients (basically you get the equations for the logarithmic derivative immediately).

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Another option which isn't geometric, but which reinforces the concept of derivative as linear approximation, is as follows.

First derive (by any means) that $\frac{\textrm{d}}{\textrm{d}u} \frac{1}{u} = -u^{-2}$.

Convey that the numerical meaning of this is $\frac{1}{u + \Delta u} \approx \frac{1}{u} - \frac{\Delta u}{u^2}$.

Use this to do some back of the envelop approximations like

$$ \begin{align*} \frac{9}{19} &= \frac{9}{20-1}\\ &\approx \frac{9}{20} - \frac{-9}{20^2}\\ &= 0.45 + 0.0225\\ &=0.4725 \end{align*} $$

Compare the approximation to the true result of $0.47368...$.

The general quotient rule repeats the same sort of calculation generally.

$$ \begin{align*} \frac{f + \Delta f}{g + \Delta g} &\approx (f + \Delta f)(\frac{1}{g} - \frac{1}{g^2}\Delta g)\\ &= \frac{f}{g} + \frac{g\Delta f - f\Delta g}{g^2} - \frac{f}{g^2} \Delta f \Delta g\\ &\approx \frac{f}{g} + \frac{g\Delta f - f\Delta g}{g^2} \end{align*} $$

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William Priestley in Calculus: A Liberal Art develops the quotient rule from the product rule and a rule for $1/f(x)$. Probably other folks do, too, but that's where I first saw it. If one follows that line of development, then one might appreciate a visualization (not a proof) for the derivative of $1/y$, akin to the change-in-rectangle one for the product rule that is fairly common that one see's echoed @Justin Hancock's answer. (The blue curve is $1/y$ vs. $y$.)

The black rectangles across the diagonal have the same area because each pair of triangles across the diagonal are congruent and the diagonal divides the rectangle in half. The area of the left rectangle is $-y\,d\left({1 \over y}\right)$, and the area of the right is ${1 \over y} \, dy$. Hence $$d\left({1 \over y}\right) = - {dy \over y^2} \,,$$ or $${d \over dx}\left({1 \over y}\right) = - {dy \big/dx \over y^2} \,.$$ Of course, one can use a finite difference $\Delta$ instead of the differential $d$ as one sees fit.

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I really appreciate the area models using differences, but here’s the kind of algebraic manipulation I enjoy, assuming the product rule in place of negative exponents:

f'= ((f/g)g)' = (f/g)g'+(f/g)'g ⇒ (f'-(f/g)g')/g = (f/g)' ⇒ (f/g)'= (f'g - fg')/(g^2)

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Multiple answers have told us the quotient rule is just (to quote one of the comments) "an algebraic reframing of the product rule."

But that misses something.

In "differential algebra" one treats derivatives only algebraically, and that's exactly what is going on when you say that starting with $$ w' = (uv)' = u'v + v'u $$ and just doing some algebra you get $$ \left(\frac wu\right)' = \frac{w'u - u'w}{u^2}. $$

However that assumes $w/u$ is differentiable if $w$ and $u$ are differentiable.

In differential algebra, everything is differentiable. In mathematical analysis, one wants to prove that this quotient, $w/u,$ is differentiable if $w$ and $u$ are differentiable.

The (full-fledged) product rule will tell us that if $w$ and $1/u$ are differentiable, then so is $w\cdot 1/u,$ i.e. $w/u.$ So the problem is to prove that if $u$ is differentiable, then so is $1/u$ (except at points where $u=0$). To that end, proceed as follows: \begin{align} \left(\frac1u\right)'(x) = {} & \lim_{\Delta x\,\to\,0} \frac{1/u(x+\Delta x) - 1/u(x)}{\Delta x} \\[8pt] = {} & -\lim_{\Delta x\,\to\,0} \frac{u(x+\Delta x) - u(x)}{ \Delta x } \cdot \frac1{u(x+\Delta x)u(x)} \\[8pt] = {} & -\lim_{\Delta x\,\to\,0} \frac{u(x+\Delta x) - u(x)}{ \Delta x } \cdot \lim_{\Delta x\,\to\,0} \frac1{u(x+\Delta x)u(x)} \\ & \text{This last “$=$” is true } \textbf{if} \text{ both } \\ & \text{limits exist and are finite.} \\ & \text{The first one exists because} \\ & \text{$u$ is differentiable. The second} \\ & \text{exists because differentiable} \\ & \text{functions are continuous and } u(x)\ne0. \\[12pt] = {} & -u'(x)\cdot \frac 1{u(x)^2} \\ & \text{and here again, the second limit} \\ & \text{is what it is because differentiable} \\ & \text{functions are continuous.} \end{align} Conclusion: The fact that $1/u$ is differentiable is not proved only by algebra plus the product rule.

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    $\begingroup$ This is true, but unfortunately does not really answer the question. $\endgroup$
    – Tommi
    Commented Jun 14 at 6:35
  • $\begingroup$ From the answer by Dave Marain: $f' = \left( \frac{f}{g} g \right)' = \left( \frac{f}{g} \right) g' + \left( \frac{f}{g} \right)' g \Rightarrow \frac{f' - \left( \frac{f}{g} \right) g'}{g} = \left( \frac{f}{g} \right)' \Rightarrow \left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2}$. The derivative of the quotient exists iff the algebraic expression involving limits $\frac{f'g - fg'}{g^2}$ exists and $g\ne0$. This algebraic combination of limits exists if $f$ and $g$ are differentiable. $\endgroup$
    – user52817
    Commented Jun 14 at 22:39
  • $\begingroup$ @user52817 : And as I said, nothing in Dave Marain's answer implies that if $f$ and $g$ are differentiable, then so is $f/g.$ Rather, it shows that the expression given by the quotient rule is the derivative IF $f/g$ is differentiable. $\endgroup$ Commented Jun 15 at 18:50
  • $\begingroup$ The equality tells us $\lim_{h\to0}\frac{(f/g)(x+h)-(f/g)(x)}{h}=\lim_{h\to0}\frac{(f(x+h)-f(x))g(x)-f(x)(g(x+h)-g(x))}{hg(x)^2}$, so the quotient is differentiable. $\endgroup$
    – user52817
    Commented Jun 16 at 17:29
  • $\begingroup$ @user52817 : But that is not in Dave Marain's answer. $\endgroup$ Commented yesterday

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