# Geometric interpretations of the quotient rule and reciprocal rule.

I have been reviewing basic calculus by working through Gilbert Strang's book "Calculus". My full time job is as a programmer, but I tutor calculus on the side. Professor Strang has kindly made the pdfs of the book, solutions manual, and study guide available for free https://ocw.mit.edu/resources/res-18-001-calculus-online-textbook-spring-2005/textbook/

I understand his 'regular' proofs of the reciprocal rule, $$\frac{d}{{dx}}\left( {\frac{1}{{v\left( x \right)}}} \right) = \frac{{ - dv/dx}}{{{v^2}}}$$

and the quotient rule, $$\frac{d}{{dx}}\left( {\frac{{u\left( x \right)}}{{v\left( x \right)}}} \right) = \frac{{v\frac{{du}}{{dx}} - u\frac{{dv}}{{dx}}}}{{{v^2}}}$$ however... I'm struggling to understand what appears to be his informal geometric derivation of both of these rules from figure 2.14, pg 74:

I know how to derive that $\frac{1}{{v + \Delta v}} - \frac{1}{v}$ is equal to $\frac{{ - \Delta v}}{{v\left( {v + \Delta v} \right)}}$, and that $\frac{{u + \Delta u}}{{v + \Delta v}} - \frac{u}{v} = \frac{{v\Delta u - u\Delta v}}{{v\left( {v + \Delta v} \right)}}$. I also know how to get from those equalities to the reciprocal rule and quotient rule. I just have not been able to figure out an intuitive way to understand the reciprocal rule and quotient rule geometrically from figure 2.14.

I found a way to use figure 2.14 and geometry to derive $\frac{1}{{v + \Delta v}} - \frac{1}{v} = \frac{{ - \Delta v}}{{v\left( {v + \Delta v} \right)}}$ and $\frac{{ - \Delta v}}{{v\left( {v + \Delta v} \right)}}$. I couldn't think of a geometric way to derive $\frac{{u + \Delta u}}{{v + \Delta v}} - \frac{u}{v} = \frac{{v\Delta u - u\Delta v}}{{v\left( {v + \Delta v} \right)}}$.

Basically, the best I was able to do for the numerator of the difference quotient for the reciprocal rule from figure 2.14 was show that a geometric derivation of that equation is consistent with the algebraic definition.

Here's the best I could come up with, using a modified version of figure 2.14 that I made in paint: For the reciprocal rule (left side of the and its top equation). $$\text{Area}_{red - triangle} = \frac{{\Delta v}}{2}$$ $$\text{Area}_{blue - triangle} = \frac{v}{2}$$ $$\text{Area}_{red - triangle} + \text{Area}_{blue - triangle} = \frac{{v + \Delta v}}{2}$$ $$v + \Delta v = 2\left( {\text{Area}_{red - triangle}} + \text{Area}_{blue - triangle} \right)$$ $$v = 2\text{Area}_{blue - triangle}$$ $$\frac{1}{{v + \Delta v}} = \frac{1}{{2\left( {Are{a_{red - triangle}} + Are{a_{blue - triangle}}} \right)}}$$ $$\frac{1}{v} = \frac{1}{{2\text{Area}_{blue - triangle}}}$$ $$\frac{1}{{v + \Delta v}} = \frac{1}{{2\left( {Are{a_{red - triangle}} + Are{a_{blue - triangle}}} \right)}}$$ $$\frac{1}{{v + \Delta v}} - \frac{1}{v} = \frac{1}{2}\left( {\frac{{ - Are{a_{red - triangle}}}}{{\left( {Are{a_{red - triangle}} + Are{a_{blue - triangle}}} \right)Are{a_{blue - triangle}}}}} \right)$$ $$\frac{1}{{v + \Delta v}} - \frac{1}{v} = \frac{1}{2}\left[ {\frac{{ - \frac{{\Delta v}}{2}}}{{\left( {\frac{{v + \Delta v}}{2}} \right)\frac{v}{2}}}} \right] = \frac{{ - \Delta v}}{{\left( {v + \Delta v} \right)\frac{v}{2}}}$$

From there... I know how to derive the reciprocal rule... taking the limit of the difference quotient as $\Delta x \rightarrow 0$ Basically seems like circular reasoning... I assume that Professor Strang included the figure to give students an intuitive, geometric feel of the quotient rule and reciprocal rule, considering that Professor Strang's informal geometric derivation of the product rule is intuitive (also... why include figure 2.14 at all if it isn't meant to help readers gain an intuitive grasp of the reciprocal and quotient rules). The pdf of his textbook is in black and white, which makes figure 2.13 on page 72 difficult to read. Here is a re-creation of figure 2.13 that I put together using paint:

And the product rule plus an informal proof of it: If $u = u(x)$ and $v = v(x)$ (assuming u and v are differentiable functions of x): $$\frac{d}{{dx}}\left( {uv} \right) = u\frac{{dv}}{{dx}} + v\frac{{du}}{{dx}}$$

Informal 'proof': $$\frac{d}{{dx}}\left( {uv} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{u\left( {x + h} \right)v\left( {x + h} \right) - u\left( x \right)v(x)}}{h}$$

As $h \to 0$,$\Delta u \to 0{\text{ and }}\Delta v \to h$

From 'Calculus' by Gilbert Strang:

The important change in area are the two strips $u \Delta v$ and $v \Delta u$. The corner area $\Delta u \Delta v$ is much smaller. When we divide by $\Delta x$ (h), the strips give $u \frac{\Delta v}{\Delta x}$ and $v \frac{\Delta u}{\Delta x}$. The corner gives $\Delta u \frac{Delta v}{\Delta x}$, which approaches zero. The extra area comes from the whole side strip.

I would greatly appreciate it if someone could help me figure out a better way to interpret figure 2.14. I know it isn't really essential for me to have an intuitive feel of the quotient rule or the reciprocal rule, however it drives me pretty crazy if I can't figure something out.

I don't know if this is great.

$v = \frac {1}{u}$

The two rectangles are equal area.

$u(-dv) = (v-dv) du\\ dv(du - u=) = \frac {1}{u} du\\ dv = -\frac {1}{u(u-du)}\ du$

Next

$v = \frac {a}{u}$

$uv = a\\ (u + du)(v+dv) = (a+da)\\ u\ dv + v\ du + dv\ du = da\\ dv(u + du) = da - \frac {a}{u}\ du\\ dv(u + du) = \frac {u\ da - a\ du}{u}\\ dv = \frac {u\ da - a\ du}{u(u + du)}$

• Thanks Doug. I don't have the time to look through your answer right now, but I'll check it out in a few hours when I have some free time. I appreciate the work you put into your answer. – Dave Rosenman Nov 28 '17 at 0:02