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Handout Daily Prep 2.5 - The Chain Rule

Section Overview

We have come a long way in just over a week’s time: from the initial basic rules for power and exponential functions, to the structure rules given by the sum and constant multiple rules, then the basic rules for the sine and cosine functions, followed by the more complicated structure rules given by the product and quotient rules. With those rules in hand, we have found derivatives for the remaining trigonometric functions, and we are now ready to deal with one more way that functions can be combined: composite functions, also known as chains of functions. The derivative rule which helps us deal with these is called the Chain Rule.

Section Basic learning objectives

These are the tasks you should be able to perform with reasonable fluency when you arrive at our next class meeting. Important new vocabulary words are indicated in italics.

State, from memory, all of the basic function and structure rules noted in the Overview above.
Identify the “inner” and ‘outer” function in a composite function of the form $C(x) = f(g(x))\text{,}$ such as in $C(x) = e^{x^2+1}$ or $C(x) = \sqrt{\cos x + 4}\text{.}$
State the Chain Rule:$\displaystyle \frac{d}{dx}f(g(x)) = \ldots$

Section To prepare for class

Complete all actions listed below. Respond to the questions highlighted with Submit.

Read motivating questions and the introduction to section 2.5 (up until Preview Activity 2.5.1).
📝 [Submit] Do Preview Activity 2.5.1.
Ask Copilot “Explain, in the simplest of terms, why the chain rule formula in calculus is the way it is. Be sure that I see the chain of variables that is involved. Maybe focus on a food chain for me.” Submit two sentences summarizing the idea returned by the AI.
Do these problems.
1. Imagine we are moving straight upward in a hot air balloon. Let $y$ be our distance from the ground. The temperature, $H\text{,}$ is changing as a function of altitude, so $H = f(y)\text{.}$ In fact, it decreases at a rate of $16^{\circ}F$ per mile. Meanwhile, our balloon is climbing at a rate of 2 miles per hour (mph). How much does our temperature change during our first 15 minutes?
2. In the previous problem, temperature is a function of height, $H = f(y)\text{,}$ and height is a function of time, $y=g(t)\text{.}$ So, we can think of temperature as a composite function of time, $H=f(g(t))$ ,with $f$ as the outside function and $g$ as the inside function. For the functions below, identify a composite function $C(t)$ by describing an outside function $f$ and an inside function $g\text{.}$
  1. $\displaystyle C(t) = (t^{2}+1)^{100}$
  2. $\displaystyle C(t) = \sqrt{3t^{2} + 5t+2}$
  3. The length $C$ (in micrometers) of steel depends on the air temperature (in $^{\circ}C$), and the temperature depends on time, $t\text{,}$ measured in hours. The length of a steel bridge increases by 0.2 micrometers for every degree increase in temperature (and is 1 micrometer at temperature of $0^{\circ}C$). The temperature is increasing at $3^{\circ}C$ per hour and is $4^{\circ}C$ at time $t=0$ hours.
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Read section 2.5.2.
Watch video Quick Review - The Chain Rule (2:20).
📝 [Submit] Do the following construction in GeoGebra that may help convince you of the accuracy of the chain rule. Submit screenshots and answers to the questions posed as needed.
- In GeoGebra, type sqrt(x) in the input box. The function $f(x) = \sqrt{x}$ will be defined and graphed.
- In the next input box, type 2*x+3. The function $g(x) = 2x+3$ will be defined and graphed.
- In the third input box, we define the composite function $h(x)=f(g(x))\text{.}$ To do so, simply type f(g(x)). GeoGebra should output $\sqrt{2x+3}\text{.}$
- We will now pick a point on the graph of $h$ to measure the slope (i.e. the derivative of $h$). In the next input box, type A = (3, h(3)). GeoGebra will plot the point $A$ at $(3,3)\text{.}$ On the fourth button (see Figure 60), choose the Tangents option. Then, click on point A on your graph followed by clicking on the graph (of $h$) itself. This will produce a tangent line to your function $h$ at $(3,3)\text{.}$ It will name this tangent line and an expression for it will appear in the next input box. From this tangent line equation, the value of $h'(3)$ should be obvious. What is it?
  
  Figure 60.
- Now we verify that $h'(3) = f'(g(3))\cdot g'(3)\text{.}$ To that end, in the next input box type g’(3). GeoGebra should compute this value and name it $a\text{.}$ Does the value returned make sense?
- Repeat this process to compute $f'(g(3))$ by typing f’(g(3)) in the next input box. GeoGebra will compute this value and name it $b\text{.}$ In the final input box, type a * b. This should be the same as $h'(3)$ (found earlier). Is it?
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Watch video Examples of the Chain Rule - Polynomials (4:32).

Checkpoint 61. 📝 [Submit] Inside and Outside.

Each composite function can be viewed as an outside function acting on an inside function. Sort the cards so that each composite function is matched with a correct description of its inside and outside functions.

The outside function is $u^5\text{.}$
The inside function is $x^2+1\text{.}$
$(x^2+1)^5$
The outside function is $\sin(u)\text{.}$
The inside function is $3x-2\text{.}$
$\sin(3x-2)$
The outside function is $e^u\text{.}$
The inside function is $x^3\text{.}$
$e^{x^3}$
The outside function is $\sqrt{u}\text{.}$
The inside function is $5x+4\text{.}$
$\sqrt{5x+4}$

Section After class

Solidifying the concepts discussed in class through practice is necessary to build your skills.

Read section 2.5.3.
Watch video Examples of the Chain Rule - Radicals (7:19).
Watch video Examples of the Chain Rule - Trigonometric Functions (5:14).
Watch video Examples of the Chain Rule - Exponential Functions (6:00).
Do Exercises 1-3 in section 2.5.6.
Watch video Chain Rule Examples - Graphs Only (5:37).
Do this problem.
1. The graph of $f(x)$ is given in Figure 62 It is known that $f'(0)=-1\text{,}$ $f'(1) = \frac{3}{4}\text{,}$ $f'(2)=1\text{,}$ and $f'(4) = -3\text{.}$ Define $h(x) = f(x^{2})\text{.}$
  
  $A coordinate graph showing a smooth curve labeled f(x). The curve decreases from the upper left, reaches a local minimum just right of x equals 0, then increases to a local maximum near x equals 3, and finally decreases again toward the lower right. The graph crosses the x-axis near x equals 1 and again near x equals 4. The x- and y-axes are labeled, with a grid and integer tick marks.$
  
  $A coordinate graph showing a smooth curve labeled h(x). The curve rises from the lower left to a local maximum left of x equals minus 2, then decreases to a local minimum near x equals minus 1. It rises slightly near x equals 0, dips again near x equals 1, and then rises to a higher local maximum near x equals 2. After that, the curve decreases steeply toward the lower right. The x- and y-axes are labeled, with a grid and integer tick marks.$
  
  Figure 62. The graph of $f(x)$ and $h(x)\text{.}$
  1. Compute $h(-1)\text{,}$ $h(0)\text{,}$ $h(1)\text{,}$ $h(\sqrt{2})\text{,}$ and $h(2)\text{.}$
  2. Compute $h'(0)\text{,}$ $h'(-2)\text{,}$ and $h'(2)\text{.}$
  3. The graph of $h(x)$ is shown in Figure 62. Do the above values make sense?
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Watch video Chain Rule Examples - Mixing Rules (10:29).
Prompt Copilot “Generate an example of a problem using a table where I would be asked to calculate the derivative of a composition of two functions using the chain rule. Be sure to also generate a solution for me.” Does the AI return a correct solution?
Read section 2.5.4.
Do Exercises 4-10 in section 2.5.6.
Start working on the MOMwork (MyOpenMath) assignment for this section.

Section Advanced learning objectives

In addition to mastering the basic objectives, here are the tasks you should be able to perform after class, with practice:

Apply the Chain Rule to basic examples.
Differentiate functions that require the application of multiple rules. For instance, compute the derivative of $\displaystyle g(x) = \frac{\sin^{2}(x) \cdot e^{x-5}}{\sqrt{x^{2}+1}}\text{.}$

Section Additional suggestions

Watch video Chain Rule - Student Problem Solving (5:53).
Watch video Average Rates of Change of Composite Functions (5:04).
Watch video How to use the Chain Rule (5:27).
Watch video Why Chain Rule Works (5:14).
Explore the following applet (especially if you are mechanical engineering major(?): The intuitive notion of the chain rule. At first glance, the moving belts will make the chain rule jump out at you!
Watch video Chain rule $u$-notation (9:27). [If you would like more examples with the chain rule...one really can’t get enough - the chain rule is a crucial topic to your success.]
Watch video Combining the Chain Rule with the Product or Quotient Rule (4:13).
Watch video Applied Example (tides) involving the differentiation of a trigonometric function (10:35).
Do these problems.
1. Consider the functions $f$ and $g$ given by the graph in Figure 63. Define $h(x) = f(g(x))\text{.}$
  
  $A coordinate graph showing two functions, f and g. Function g is drawn as a solid blue piecewise linear graph: it rises steeply from the lower left to a peak near x equals 2, then decreases steadily to the right. Function f is drawn as a red dashed curve: it decreases to a minimum near x equals 1, then increases and continues upward to the right. The two graphs intersect at a point near x equals 4. The x- and y-axes are labeled, with a grid and integer tick marks.$
  
  Figure 63. The graph of $f(x)$ and $g(x)\text{.}$
  1. Compute $h'(1)\text{.}$
  2. Compute $h'(0)\text{.}$
  3. Does $h'(2)$ exist? If so, compute its value. If not, why not?
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2. Use information from the graphs of $f(x)$ and $g(x)$ and the tangent lines shown in Figure 64 to calculate $m'(2)$ if $m(x) = g(f(x))\text{.}$
  
  $A coordinate graph showing a smooth blue curve labeled f(x). A filled point on the curve is labeled with coordinates (2, 3). From that point, a straight black line extends upward to the right to a filled point labeled (4, 4.8). The straight segment lies above the curve between the two points. No axes or grid are shown.$
  
  $A coordinate graph showing a smooth blue curve labeled g(x). A horizontal segment at y equals 5 has filled endpoints labeled (0, 5) and (2, 5). At x equals 2, the curve reaches a local maximum at the filled point (2, 5) and then decreases to the right. Two filled points on the decreasing part of the curve are labeled (3, 4) and (3.5, 3). A straight black line segment is tangent to the curve near the point (3.5, 3).$
  
  Figure 64. The graphs of $f(x)$ and $g(x)$ and some tangent lines.
3. If $g(x) = \sqrt{f(x)}\text{,}$ where the graph of $f$ and a tangent line is shown in Figure 65, evaluate $g'(3)\text{.}$
  
  $A coordinate graph showing two functions. A solid blue straight line decreases from left to right, starting above y equals 4 on the left and ending below y equals 1 near x equals 6. A red dashed curve increases from the left to a peak near x equals 2, then decreases and approaches the blue line. The two graphs meet at a filled point near x equals 3. Past x equals 6, the red dashed curve turns upward slightly while the blue line continues downward. The x- and y-axes are labeled, with a grid and integer tick marks.$
  
  Figure 65. The graph of $f(x)$ and a tangent line at $x=3\text{.}$
4. For each of the following functions of $x\text{,}$ write the equation for the derivative function. Please do us both a favor and don’t simplify the answers.
  1. $\displaystyle y = \sin 3x$
  2. $\displaystyle w = (\sin 3x)^{3}$
  3. $\displaystyle u = (\sin 3x)^{3} + 5x$
  4. $\displaystyle v = \left[ (\sin 3x)^{3} + 5x \right]^{2}$
  5. $\displaystyle \displaystyle k(x) = x + \frac{1}{x}$
  6. $\displaystyle \displaystyle l(x) = \sqrt{x + \frac{1}{x}}$
  7. $\displaystyle \displaystyle m(x) = \left[ (\sin 3x)^{3} + 5x \right]^{2} \cdot \sqrt{x + \frac{1}{x}}$
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5. Differentiate $\displaystyle y = \sqrt{1+e^{\sqrt{3+x^2}}}\text{.}$
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Section Answers

Subsection To prepare for class

$\frac{dH}{dy}= -16\frac{^{\circ}F}{mi}$ and $\frac{dy}{dt}= 2 \frac{mi}{hr}$ means that $\frac{dH}{dt}= \frac{dH}{dy}\cdot \frac{dy}{dt}= (-16)(2) = -32 \frac{^{\circ}F}{hr}\text{.}$ In 15 minutes, $H$ changes $1/4$ of this or $-8^{\circ}F$ (a decrease of 8 degrees Fahrenheit).
1. $g(t) = t^{2}+1$ and $f(t) = t^{1}00\text{,}$ though answers may vary.
2. $g(t) =3t^{2} + 5t + 2$ and $f(t) = \sqrt{t}\text{,}$ though answers may vary.
3. $g(t) =3t+4$ and $f(t) = 0.2t+1\text{,}$ though answers may vary.
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Subsection After class

1. $h(-1)=f(1)=0\text{,}$ $h(0)=f(0)=0\text{,}$ $h(1)=f(1)=0\text{,}$ $h(\sqrt{2})=f(2)=1\text{,}$ $h(2)=f(4)=0\text{.}$
2. $h'(x)=f'(x^{2})\cdot 2x$ so that $h'(0)=f'(0)\cdot (2)(0)=0\text{,}$ $h'(-2) = f'(4)\cdot (2)(-2) = 12\text{,}$ and $h'(2)=f'(4)\cdot (2)(2) = -12\text{.}$
3. Yes
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Subsection Additional suggestions

1. $h'(x) = f'(g(x))g'(x)$ so that $h'(1) = f'(g(1))g'(1) = f'(3)g'(1) = f'(3)\cdot 3 = \frac{2}{3}(2)(3) = 4\text{;}$ Note that on $(-2,4)\text{,}$ the function $f(x)$ can be found to be $f(x) = \frac{1}{3}(x-1)^{2} + 1$ by noting it has vertex at $(1,1)$ and the point $(4,4)$ determines $c$ in $f(x) = c(x-1)^{2} + 1\text{.}$
2. $h'(0)=f'(g(0))g'(0)=f'(0)\cdot 3 = \frac{2}{3}\cdot 3 = 2\text{.}$
3. $h'(2)$ does not exist since $g'(2)$ does not exist. $h$ is not ‘smooth‘ at $x=2\text{.}$
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$m'(x) = g'(f(x))f'(x)$ so that $m'(2) = g'(f(2)) \cdot f'(2) = g'(3) \cdot \left( \frac{4.8-3}{4-2}\right) = \left( \frac{4-3}{3-3.5}\right) (0.9) = -2(0.9) = -1.8\text{.}$
$\displaystyle g'(x) = \frac{1}{2}f(x)^{-1/2}\cdot f'(x) = \frac{1}{2\sqrt{f(x)}}\cdot f'(x)\text{.}$ Thus, $\displaystyle g'(3) = \frac{1}{2\sqrt{f(3)}}\cdot f'(3) = \frac{1}{2\sqrt{2}}\cdot \left( \frac{2-4}{3-0}\right) = -\frac{1}{3\sqrt{2}}\text{.}$
1. If $z=3x\text{,}$ then $\displaystyle \frac{dy}{dx}= \frac{dy}{dz}\cdot \frac{dz}{dx}= \cos z \cdot 3 = 3 \cos(3x)\text{.}$
2. Here, $w= y^{3}$ so that $\displaystyle \frac{dw}{dx}= \frac{dw}{dy}\frac{dy}{dz}\frac{dz}{dx}= 3y^{2} \cdot \cos z \cdot 3 = 9 \sin^{2} (3x) \cos (3x)\text{.}$
3. Here $u = w^{3} + 5x$ so that $\displaystyle \frac{du}{dx}= \frac{du}{dw}\frac{dw}{dx}+ 5 = 3w^{2} \cdot 9 \sin^{2} (3x) \cos (3x) + 5 = 27 \sin^{6} (3x) \cdot \sin^{2} (3x) \cos (3x) + 5 = 27 \sin^{8} (3x) \cos (3x) + 5\text{.}$
4. Here $v = u^{2}$ so that $\displaystyle \frac{dv}{dx}= \frac{dv}{du}\frac{du}{dx}= 2u \frac{du}{dx}= 2 \left( \sin^{3} (3x) + 5x \right) \left( 27 \sin^{8} (3x) \cos (3x) + 5 \right)\text{.}$
5. $\displaystyle \frac{dk}{dx}= 1 -\frac{1}{x^{2}}\text{.}$
6. Here, $l = \sqrt{k}$ so that $\displaystyle \frac{dl}{dx}= \frac{dl}{dk}\frac{dk}{dx}= \frac{1}{2}k^{-1/2}\left( 1-\frac{1}{x^{2}}\right) = \frac{1}{2\sqrt{x+\frac{1}{x}}}\left(1 - \frac{1}{x^{2}}\right)\text{.}$
7. Here $m(x) = v(x) \cdot l(x)$ so that by the product rule $\displaystyle \frac{dm}{dx}= \frac{dv}{dx}l(x) + v(x) \frac{dl}{dx}$ =
  
  \begin{equation*} 2 \left( \sin^{3} (3x) + 5x \right)(27 \sin^{8} (3x) \cos (3x) + 5 ) \sqrt{x + \frac{1}{x}}+ \left( \sin^{3} (3x) + 5x \right)^{2} \cdot \frac{1}{2\sqrt{x+\frac{1}{x}}}\left(1 - \frac{1}{x^{2}}\right). \end{equation*}
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$\displaystyle \frac{dy}{dx}= \frac{1}{2}\left( 1+e^{\sqrt{3+x^{2}}}\right)^{-1/2}\cdot e^{\sqrt{3+x^{2}}}\cdot \frac{1}{2}(3+x^{2})^{-1/2}\cdot 2x\text{.}$

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