A visualization for the quotient rule

Question

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?

Answer 1

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

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

Answer 2

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.)

Answer 3

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).

Answer 4

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.

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

Answer 5

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.

Answer 6

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)

Answer 7

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.