A visualization for the quotient rule

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?

• I'm curious what grades “middle school” refers to in Norway. In America this is something students would encounter at the very end of secondary education, if ever. Apr 17 at 5:45
• Middle school (ungdomsskole) is grades 8-10. This education allows one to teach grades 5-10. We wish the teachers to know more mathematics than their pupils do. The pupils do not meet derivatives in ungdomsskole. Apr 17 at 9:01
• 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. Apr 17 at 18:27
• @adam.baker my school district had "accelerated learning classes" where they did teach some basic calculus stuff at 6th and 7th grade, but only a few students qualified.
– qwr
Apr 19 at 15:00
• 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.
– qwr
Apr 19 at 15:02

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*}

• This is a clever and nice answer. It brings to attention that the quotient rule is just an algebraic reframing of the product rule. Apr 16 at 15:22

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 $$\Delta y = \frac{\Delta A}{x}$$.

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.

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.

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.

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

(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.)

• 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 . Apr 17 at 17:56
• @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. Apr 18 at 2:23

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).

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*}

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)