Calculus I Lesson 10: Chain Rule

  1. Warm Up
  2. Final Paper
    1. Prompt 1: Calculus in the news
    2. History
    3. Topic Statement
    4. Other deadlines
  3. Problem Presentations
  4. Recap:
    1. Derivative of $\sec(x)$
  5. Chain Rule
    1. Example
    2. Exercises
    3. Solutions
  6. Reminders

Warm Up

Use the quotient rule to find the derivatives of the following functions:

  1. $f(x) = \frac{1}{x}$.
  2. $g(x) = \frac{1}{x^2}$.
  3. $\sec(x)$. Recall that $\sec(x) = \frac{1}{\cos(x)}$.

Final Paper

Prompt 1: Calculus in the news

“In the fall of 1972 President Nixon announced that the rate of increase of inflation was decreasing. This was the first time a sitting president used the third derivative to advance his case for reelection.”

Hugo Rossi, Notices of the American Mathematical Society (October 1996)

“The rate of increase is slowing. But the number of cases are still going up.”

(then)-Governor Cuomo (March 27, 2020)

Both of these quotes are discussing how a quantity is changing over time: that is, they are all echoing the concept of the derivative. In fact: both refer to higher order derivatives.

Prompt 1: Find an article / news clip / podcast / etc in which two quantitative variables are related.

History

Prompt 2: Find out as much as you can about any mathematician who contributed to the development of calculus and describe their contributions as deeply as you can. This paper can be either more historical, or mathematical.

If you’re interested in mathematical topics?

If you’re interested in historical topics?

Either way:

Some mathematicians of interest:

Many others. Do some initial research on history of calculus and see what piques your interest!

Topic Statement

Due October 20. Pick a specific topic you’d like to study. Narrow it down as much as you can, and give a short summary of what you’ll study.

Other deadlines

Problem Presentations

You will have two problem presentations this semester. The idea:

Keep the presentation short: about 5 minutes or so in total.

You will be evaluated (out of 5 points) on the following criteria:

You can choose to present the problem on the chalkboard / whiteboard or you can create slides (PPT, Keynote, Google Slides, whatever) and upload them to the assignment space on BrightSpace.

We will be presenting on the following dates:

I can take volunteers for specific dates or I can randomly assign students to present.

Recap:

Recall: the quotient rule

\[(\frac{f(x)}{g(x)})^\prime = \frac{f^\prime(x)g(x) - g^\prime(x)f(x)}{(g(x))^2}\]

and the trig derivatives:

Last time, we looked at $f(x) = \cos(x)$. In particular, $f^\prime(0) = 0$. Why is this? Look at the unit circle. Remember that $\cos(\theta)$ is the $x$-value of the point on the unit circle at angle $\theta$.

Unit Circle\

Near $\theta = 0$, how quickly does that $x$-value change? What about near $\theta = \frac{\pi}{2}$?

Derivative of $\sec(x)$

$\sec(x) = \dfrac{1}{\cos(x)}$. So again use the quotient rule:

\[\dfrac{0 - (-\sin(x))}{(\cos(x))^2}\]

We can write this as $\dfrac{1}{\cos(x)} \dfrac{\sin(x)}{\cos(x)} = \sec(x)\tan(x)$.

Chain Rule

Recall function composition:

Theorem: The Chain Rule. Suppose $f$ and $g$ are differentiable. Then $(f \circ g)^\prime(x) = f^\prime(g(x)) g^\prime(x)$.

Why? How does this end up happening? Let’s do an example, without using the chain rule, and see if we can “spot” it showing up.

Let $f(x) = (\sin(x))^2$. Let’s try to compute the derivative at $x = \pi$. First we will need to look at the following rate of change, and then consider the limit at $x \rightarrow \pi$:

\[\frac{f(x) - f(\pi)}{x - \pi} = \frac{(\sin(x))^2 - (\sin(\pi))^2}{x - \pi}\]

Instead of computing this directly right now, notice that as $x \rightarrow \pi$, there is some change in $\sin(x)$, which corresponds to some change in $(\sin(x))^2$. How do we get the derivative of $\sin(x)$ to show up in the above fraction? Multiply by $\frac{\sin(x) - \sin(\pi)}{\sin(x) - \sin(\pi)}$!

\[\frac{(\sin(x))^2 - (\sin(\pi))^2}{x - \pi} \cdot \frac{\sin(x) - \sin(\pi)}{\sin(x) - \sin(\pi)} = \frac{(\sin(x))^2 - (\sin(\pi))^2}{\sin(x) - \sin(\pi)} \cdot \frac{\sin(x) - \sin(\pi)}{x - \pi}\]

Let’s look at each of these fractions individually. One of them, as $x \rightarrow \pi$, should be obvious: $\frac{\sin(x) - \sin(\pi)}{x - \pi} \rightarrow \cos(\pi)$, since this is just the derivative of $\sin(x)$ (evaluated at $x = \pi)$.

The other one: $\frac{(\sin(x))^2 - (\sin(\pi))^2}{\sin(x) - \sin(\pi)}$ is interesting. Let’s make a change of variables. Let $u = \sin(x)$. Then as $x \rightarrow \pi$, $u \rightarrow \sin(\pi)$. Now this looks like: \(\frac{u^2 - (\sin(\pi))^2}{u - \sin(\pi)},\) which, as $u \rightarrow \sin(\pi)$, is just the derivative of $u^2$ evaluated at $\sin(\pi)$. In other words, it’s $2\sin(\pi)$.

So the whole thing is $2 \sin(\pi) \cos(\pi)$! Since $\sin(\pi)$ is zero, this is actually just 0.

Example

Let $f(x) = \sin(x^2)$. Then $f^\prime(x)$? How do we figure this out?

So our final answer: $(\sin(x^2))^\prime = 2x \cos(x^2)$

Exercises

Find the derivatives of the following:

  1. $f(x) = \sin(2x)$
  2. $h(x) = (\cos(x))^2$

Solutions

Check your answers:
  1. $(\sin(2x))^\prime = 2\cos(2x)$
  2. $(\cos(x))^2 = -2\sin(x)\cos(x)$

Reminders