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# Section 6.3 A Garden of Exact Rate of Change Functions

## 6.3.1 Motivation

Here’s a secret: GC uses approximations to compute $\int_a^x r_f(t)dt$ for any function $r_f$. When GC integrates a function over a large domain it can be quite inaccurate because it always uses 512 subintervals. So when GC computes $$\int_0^{1024} r_f(t)dt$$ it uses subintervals of size 2, which could produce a highly inaccurate result if $r_f$ varies dramatically within those subintervals.

However, section 6.2 gave us the insight that we can compute an integral of a function $r_f$ exactly whenever we know an accumulation function f in closed form that has $r_f$ as its rate of change function. This fact bears repeating.

 Whenever we know a closed form representation of a function f that that has $r_f$ as its exact rate of change function, we can use the closed form representation of f to compute $\int_a^x r_f(t)dt$ exactly, namely $$\int_a^x r_f(t)dt=f(x)-f(a).$$We do not need to approximate the integral.

This fact motivates us to derive rate of change functions in closed form from accumulation functions in closed form. As we derive more accumulation-rate pairs in closed form, our ability to  compute integrals exactly will grow.

A general remark will be helpful in all the remaining sections. It is that when $0<h<1$ and n is a positive integer, $h^{n+k}$ is smaller than $h^n$ by a factor of $h^k.$ If $h=0.01,$ then $0.01^{15}$ is smaller than $0.01^{10}$ by a factor of $0.01^5,$ or $\left(\frac{1}{100}\right)^5.$ Stated generally, for $0<h<1$, as $h$ becomes small, $h^n$ becomes smaller rapidly as the value of n increases.

### Practice

Here is a sheet of practice problems. Examine it repeatedly during your study of this chapter for problems that you can solve with methods you have learned up to the moment that you are examining the sheet. You will learn a method for checking your work graphically. Use it.

## 6.3.2 Rate of Change of a Constant Function and a Constant Times a Function

Intuitively, if a quantity has the same measure at every moment of its independent variable, then its value is not changing and its rate of change is 0. More formally,

 $f(x)$ $=c$ $r(x)$ $=\dfrac{f(x+h)-f(x)}{h}$ $= \dfrac{c-c}{h}$ $= 0$ $r_f(x)$ $= 0$

Equation 6.3.1.

In regard to recovering f from its rate of change function, we have the seemingly odd result that when $f(x)=c$, $\int_a^x r_f(t)dt=0$, not c. This oddness is only apparent. If you think of f as an accumulation function and $\int_a^x r_f(t)dt$ as f’s accumulation from a to x, nothing additional has accumulated. Put another way, we know that $f(x) = c$ for some constant c when $r_f(x)=0$ for all x, so by the FTC, $\int_a^x 0dt = f(x)-f(a) = c-c=0$.

When $g(x)=c \cdot f(x)$, where c is a constant, then
\begin{align} r_g(x) & = \frac{g(x+h)-g(x)}{h} \\ & = \frac{c \cdot (f(x+h)-f(x))}{h} \\ & = c \cdot \frac {f(x+h)-f(x)} {h} \\ & = c \cdot r_f(x) \\ \end{align}
Therefore, by the FTC, $\int_a^x c r_f(t)dt = c \int_a^x r_f(t)dt = c (f(x)-f(a))$

## 6.3.3 Rate Functions for Power Functions

Let $f(x)=cx^n$, n a positive integer and $c≠0$. We want $r_f$, f's exact rate of change function, in closed form. We’ll start with $n=3$ and use that to generalize to arbitrary integer values of $n≥0$. We’ll also start with the approximate rate of change function and get the exact rate of change function by making the value of h so small that $h \doteq 0$. If our derivation is correct, then the graphs of the open form and closed form rate of change functions should coincide, and the graphs of the open form and closed form integrals should coincide.

### Example:

 $f(x)$ $=4x^3$ $r(x)$ $=\dfrac{f(x+h)-f(x)}{h}$ $=\dfrac{4(x+h)^3-4x^3}{h}$ $=\dfrac{4(x^3+3x^2h+3xh^2+h^3)-4x^3}{h}$ $=\dfrac{4 \cdot 3x^2h+4 \cdot 3xh^2 + 4h^3}{h}$ $=4 \cdot 3x^2+4 \cdot 3xh + 4h^2$ $r_f(x)$ $\doteq 4 \cdot 3x^2$ when $h \doteq 0$

### Generalization:

 $f(x)$ $=cx^n, c \neq 0, n \ge 0$ $r(x)$ $=\dfrac{f(x+h)-f(x)}{h}$ $=\dfrac {c(x+h)^n - cx^n}{h}$ $= \dfrac{c(x^n + nx^{n-1}h + n(n-1)x^{n-2}h^2 + ... + nxh^{n-1} + h^n) - cx^n}{h}$ $= \dfrac{cnx^{n-1}h + cn(n-1)x^{n-2}h^2 + ... + cnxh^{n-1} + ch^n)}{h}$ $= cnx^{n-1} + cn(n-1)x^{n-2}h + ... + cnxh^{n-2} + ch^{n-1}$ $r_f(x)$ $\doteq cnx^{n-1}$ when $h \doteq 0$

### Check Example:

Let $f(x)=4x^3$. Use GC to see whether the graph of $y=12x^2$ coincides with the graph of $y=\dfrac{f(x+h)-f(x)}{h}$ when $h=0.0001$. What does this tell you?

Use GC to see whether the graphs of $y=\int_a^x 12t^2dt$ and $y=4x^3-4a^3$ coincide. What does this tell you?

### Check Generalization:

Let $f(x)=cx^n$. Use GC to see whether the graph of $y = cnx^{n-1}$ coincides with the graph of $y=\frac{f(x+h)-f(x)}{h}$ when $h=0.0001$ and n is any arbitrary value. What does this tell  you?

Use GC to see whether the graphs of  $y=\int_a^x cnt^{n-1}dt$ and $y=c(x^n-a^n)$ coincide for arbitrary values of c and n. What does this tell you? In what way is this result an instance of the FTC?

Library of ROC/Integral pairs, Version 1

 Accum Fn: $f(x)=$ Rate of Change: $r_f(x)=$ Accum Fn from Rate: ${\displaystyle \int_a^x r_f(t)dt}$ c 0 0 $cx^n$ $cnx^{n-1}$ $cx^n-ca^n$

### Applications

1) Let $g(x)=4$. What is $\int_a^x g(t)dt$?

Rewrite g as $g(x)=4 \cdot 1x^0$. Then $\int_a^x g(t)dt = \int_a^x 4 \cdot 1t^0dt = 4(x^1-a^1)$.

Reflection 6.3.1. Think of what $g(x)=4$ means when g is a rate of change function. Give a concrete example that illustrates how it make sense that $\int_a^x 4dt = 4(x-a)$.

2) Let $g(x)=7x^3$. What is $\int_a^x g(t)dt$ in closed form?

Were $g$ defined as $g(x)=4x^3$ we would know by the FTC that $\int_a^x 4t^3 dt = x^4 - a^4$. But the coefficient of $x^3$ is 7, not 4. We can adjust for this by noting that $7 = \frac{7}{4} \cdot 4$, so we can write $g(x)=7x^3$ as $g(x)=\frac{7}{4}(4t^3)dt$, so $\int_a^x 7t^3dt = \int_a^x \frac{7}{4}(4t^3)dt=\frac{7}{4}(x^4-a^4)$.

## 6.3.4 Rate Functions for Sum Functions

If you walk away from your friend on a straight sidewalk at 6 $\mathrm{\frac{km}{hr}}$ and your friend walks away from you at 4 $\mathrm{\frac{km}{hr}}$.  Every hour that the two of you walk, you will get 6 km further from him and he will get 4 km further from you. So, the distance between the two of you increases at a rate of $(6 + 4)$ $\mathrm{\frac{km}{hr}}$.

In general, if $k(x) = f(x)+g(x)$ and if $r_f$ and $r_g$ exist, then $r_k(x) = r_f(x)+r_g(x)$. We can prove the general statement as follows. Assume that $k(x) = f(x)+g(x)$ and that $r_f$ and $r_g$ exist for all values of x. Then,

 $k(x)$ $=f(x)+g(x)$ $r(x)$ $=\dfrac{k(x+h)-k(x)}{h}$ $=\dfrac{(f(x+h)+g(x+h))-(f(x)+g(x))}{h}$ $= \dfrac{(f(x+h)-f(x))+(g(x+h)-g(x))}{h}$ $= \dfrac{(f(x-h)-f(x))}{h} + \dfrac{(g(x-h)-g(x))}{h}$ $r_k(x)$ $\doteq r_f(x) + r_g(x)$ when $h \doteq 0$

Equation 6.3.2.

Library of ROC/Integral pairs, Version 2

 Accum Fn: $f(x)=$ Rate of Change: $r_f(x)=$ Accum Fn from Rate: ${\displaystyle \int_a^x r_f(t)dt}$ c 0 0 $cx^n$ $cnx^{n-1}$ $cx^n-ca^n$ $q(x)+p(x)$ $r_q(x)+r_p(x)$ $(q(x)+p(x))-(q(a)+p(a))$

### Application

Let $h(x)=5x^2-3x+4$. What is $r_h(x)$? What is $\int_a^x h(t)dt$ in closed form?

We can rewrite h as $h(x)=5x^2+(-3x)+4$. Then to find $r_h(x)$:

 Let $h(x)$ $= f(x)+g(x)+k(x)$ where $f(x)=5x^2$, $g(x)=-3x$, and $k(x)=4$ $r_h(x)$ $=r_f(x) + r_g(x) + r_k(x)$ $= 10x + -3 + 0$ $= 10x-3$

$\text{Let }h(x) = f(x)+g(x)+k(x) \text{ where }f(x)=5x^2, g(x)=-3x^1\text{, and }k(x)=4x^0.$

We can use the FTC to find $\int_a^x h(t)dt$ in closed form:
\begin{align} \int_a^x h(t)dt &= \int_a^x f(t)dt + \int_a^x g(t)dt +\int_a^x k(t)dt\\[1ex] &= \int_a^x 5t^2dt + \int_a^x -3t^1dt +\int_a^x 4t^0dt\\[1ex] &= \int_a^x \frac{5}{3} \cdot 3t^2dt + \int_a^x -\frac{3}{2} \cdot 2t^1dt +\int_a^x 4t^0dt\\[1ex] &= \frac{5}{3} \int_a^x 3t^2dt - \frac{3}{2}\int_a^x 2t^1dt +4\int_a^x t^0dt\\[1ex] &=\frac{5}{3}(x^3-a^3)-\frac{3}{2}(x^2-a^2)+4(x-a)\end{align}

Notice in this last example that we changed, for example, $\int_a^x5t^2dt$ to the equivalent statement $\int_a^x \frac{5}{3}\cdot 3t^2dt$. We then rewrote the second statement as $\frac{5}{3}\int_a^x 3t^2dt$. This sleight of hand was simply to put the integral into a form that is easier to apply the FTC, namely that $\int_a^x 3t^2dt=x^3-a^3$.

## Exercise Set 6.3.4

1. Given that $f(x)=\dfrac{1}{2}x^3-2x+4$, use $r(x)=\dfrac{f(x+h)-f(x)}{h}$ to derive f’s exact rate of change function, $r_f$, expressed in closed form. Check your derivation by graphing it along with the graph of $y=r(x)$, with $h=0.001$.

2. Find the exact rate of change function, $r_h$, in closed form for each function h defined below. Use GC to check your answers by graphing $y=r(x)$ and $y=r_h(x)$.

1. $h(x)=5x^4-10x+2$

2. $h(x)=\dfrac{5}{3}x^2-1$

3. $h(x)=\dfrac{1}{4}x^6 + \dfrac{1}{3}x^3 - x^4$

3. Write $\int_a^x h(t)dt$ in closed form for each function h in Exercise 2. Use GC to check your answers.

4. In one kind of chemical reaction two separate substances combine to form a third substance.  The function g, defined as $g(t)=18t-3t^2$, gives the accumulated amount of the third substance t minutes since the reaction began.

1. What is $r_g$, g’s exact rate of change function, in closed form?

2. At what rate is the third substance being produced at the moment that 2 minutes have passed?  At the moment that 4 minutes have passed?

3. Describe in your own words what (b) tells you about the production of the third substance at the 2-minute and 4-minute mark. (adapted from Kline, p. 35)

5. Angry Birds is a popular game for mobile devices.  The goal of Angry Birds is to use a slingshot to shoot angry birds at green pigs.  This goal is achieved by knocking down walls that protect the pigs.  The path of a red bird that is fired from a slingshot can be modeled by the function h, defined as $h(t)=-16t^2+48t+8$, where $h(t)$ is the height of the bird (in meters) off the ground, and t is the number of seconds that have elapsed since the slingshot was released.

1. What is the exact rate of change function, $r_h$, whose values give the rate at which the red bird’s height changes at each moment in time?

2. What restrictions should we place on our dependent and independent variables?  Why?

3. What is the maximum height that the red bird will reach?  Use your rate of change function to justify this solution.

4. Blue birds fly into the screen from the left side. We know that a blue bird’s rate of change of distance from the ground at any moment is modeled by $r_f(t)=-32t+20$ m/s, where t is the number of seconds since it entered the screen.  Determine the blue bird’s change in height between 0.5 seconds and 1 seconds assuming that it entered the screen 10 meters above the ground.

5. Compare the blue bird’s maximum height to the red bird’s maximum height.

6. The binomial coefficients, denoted ${n \choose k}$ where $${n \choose k} = \frac{n!}{(n-k)!k!},$$ are the coefficients of the powers of x when $(x+1)^n$ is expanded.  That is,
\begin{align} (x+1)^n & = {n \choose 0}x^0 \cdot 1^n + {n \choose 1}x^1 \cdot 1^{n-1} + {n \choose 2}x^2 \cdot 1^{n-2} + ... + {n \choose {n-1}} x^{n-1} \cdot 1^1 + {n \choose n}x^n \cdot 1^0 \\ & = {n \choose 0}x^0 + {n \choose 1}x^1 + {n \choose 2}x^2 + ... + {n \choose {n-1}} x^{n-1} + {n \choose n}x^n \\ \end{align}

In Section 6.2 we found that when $f(x)=(x+1)^n, r_f(x) = n(x+1)^{n-1}.$ Use the expansion of $(x+1)^n$ given above to find another representation of the exact rate of change of $f(x)=(x+1)^n$ with respect to x. (Hint: Keep in mind that all terms of the form ${n \choose k}$ are constants in the expansion of $(x+1)^n$).

7. Explain why it is reasonable that $f(x)=x^n+a$ and $g(x)=x^n$ should have the same rate exact rate of change function. (Hint: Consider the relationship between the graphs of f and g.)

8. In Section 3.13, Exercise 5, we presented a scenario where two baseball players were running from first to second base and from second to third base, respectively. Player 1 ran at 18 $\mathrm{\frac{ft}{sec}}$.

1. What does $\int_0^5 18dt$ represent in this situation?  Evaluate this integral as part of your explanation.

2. Derive the exact rate of change function for $\int_a^x 18dt$ and explain what it means in regard to Player 1.

## 6.3.5 Rate Functions for Composite Functions

Suppose that k is a composite function such that $k(x)=g(f(x))$. It seems intuitive to think that $r_k$, the exact rate of change function for k, would be $r_k(x)=r_g(f(x))$. Were this true, then if $k(x)=(e^x)^2$, $r_k$ would be $r_k(x)=2(e^x)^1$. Figure 6.3.1 shows that our intuition is wrong. The graph of r, which we know should be a reasonably good approximation to k’s exact rate of change function, is not even close to the graph of $y=2e^x$. So $r_k(x)≠2e^x$ when $k(x)=(e^x)^2$.

Figure 6.3.1. Testing our intuition about the rate function for a composite function.

### Rate of change magnifies changes

A useful way to think about the composite function k where $k(x)=g(f(x))$ is to remind ourselves that $f(x)$ is the argument to g, not g’s independent variable. If we were to say $u=f(x)$ and ask about the rate of change of g with respect to u, then $r_k(u)$ would indeed be 2u. But u is changing at a rate of its own with respect to x, so it is as if changes in x get magnified by f, and then changes in f get magnified by g.

Let’s go further with thinking of rate of change as magnification. Suppose you have two lenses, f and g, and that f magnifies objects to make them appear 2 times as large, and g magnifies objects to make them appear 3 times as large. Think of holding g over a rubber stick. The stick, to your eye, is magnified 3 times its original length. Now slip lens f between lens g and the stick. Lens f magnifies the stick by a factor of 2. Lens g magnifies the magnified image it receives from lens f by a factor of 3. So in the image you see, the stick has been magnified twice, so its total magnification is by a factor of $3 \cdot 2$.

Here is where the idea of rate of change as magnification comes in. Call the length of the stick x. Now imagine that you stretch the stick to make it a little longer—dx longer. Lens f magnifies that change so that the change of dx in x is magnified in f’s image to be 2dx. The rate of change of f’s image with respect to changes in x is 2. Lens g magnifies the magnified change by a factor of 3, so in the final image the change of dx in the stick’s length is magnified to be twice as large by f and then three times as large by g. The change of dx in x becomes a change of $3 \cdot 2 \mathrm{d}x$ in the image produced by g. That is, if we call the length of the the stick in g’s image y, then $\mathrm{d}y=3 \cdot 2 \mathrm{d}x$. The rate of change of the image produced by $g(f(x))$ is 6 with respect to changes in x.

So it is with functions in general: Changes in x are magnified by $r_f$, and changes in f are magnified by $r_g$. When $k(x)=g(f(x))$, we have $r_k(x)=r_g(f(x))r_f(x)$. This chain of magnifications is shown in Figure 6.3.2.

Note that Figure 6.3.2 shows three number lines, vertically. The first is the x number line, the second is the f(x) number line, and the third is the k(x) = g(f(x)) number line. In all three, up is positive and down is negative.

Figure 6.3.2 also shows changes in each of x, f(x), and g(f(x)). The arrows show how the changes are related.

Figure 6.3.2. dx is magnified by $r_f(x)$ to make df, then df is magnified by $r_g(f(x))$ to make dk.

Figure 6.3.2 shows two views of the chain of changes in x, f(x), and g(f(x). The first view shows the sequence of changes. Changes in x (i.e., dx) are magnified by $r_f(x)$ to get a change in f (df), and then df is magnified by $r_g(f(x))$ to get dk. Keep in mind that magnifications can be greater than 1 or less than 1, and that magnifications can be negative (flipped image). Also keep in mind that it is changes that are magnified, not the value of x or the value of $f(x)$.

The second view shows directly how changes in x are related to changes in k(x) = g(f(x)).

The equation $r_k(x)=r_g(f(x))r_f(x)$, being a chain of magnifications, is called the chain rule.

Reflection 6.3.2 Move the movie slider in Figure 6.3.2 to 9 seconds. Answer each of the following:

• Is dx positive or negative? Explain what this means.
• Is df positive or negative? Explain what this means.
• Is dk positive or negative? Explain what this means.
• Is $r_f$ positive or negative? Explain what this means.
• Is $r_g(f(x))$ positive or negative? Explain what this means.

### Application

Let the functions j, k, and s be defined as $j(u)=u^2+1$, $k(x)=x^3$, and $s(t)=k(j(t))$. According to Version 2 of our Library, $r_j(u)=2u$ and $r_k(x)=3x^2$. Therefore, by the general rule that $r_s(x)=r_k(j(x))r_j(x)$, we have that $r_s(x)=(3(x^2+1)^2)(2x)$.

Let’s check this in GC. Figure 6.3.3 starts with the closed form accumulation functions $j(u)=u^2+1$, $k(x)=x^3$, and $s(t)=k(j(t))$. It also shows GC’s graph of s’s approximate rate of change function, r. Click on the right arrow to see that GC’s graph of our derived function $r_f$ coincides with the graph of r. Since we trust GC’s graph of r to be a close approximation of s’s actual rate of change function’s graph, Figure 6.3.3’s second image supports our analysis.

 (a) (b)

Figure 6.3.3. Testing ROC of composite function. Scroll between first and second image.

### Application: Using the Chain Rule to extend the Power Rule

We derived the general rate of change function for a power function whose exponent is a natural number, namely $r_f(x)=nx^{n-1}$ when $f(x)=x^n$. We will use the chain rule together with the fact that $x^{-n}=\left(x^{-1}\right)^n$ to extend the power rule to the case of any accumulation function that is raised to a negative integer power. To do this we must first derive the rate of change function for $f(x)=x^{-1}$, or $f(x)=\frac{1}{x}$. We do this in Equation 6.3.2.
\begin{align} f(x)&=\frac{1}{x}\\[1ex] r(x)&=\frac{f(x+h)-f(x)}{h}\\[1ex] &=\frac{\left(\dfrac{1}{x+h}\right)-\left(\dfrac{1}{x}\right)}{h}\\[1ex] &=\frac{\dfrac{x-(x+h)}{x(x+h)}}{h}\\[1ex] &=\frac{x-x-h}{hx(x+h)}\\[1ex] &=\frac{-h}{hx(x+h)}\\[1ex] &=\frac{-1}{x(x+h)}\\[1ex] r_f(x)&\doteq \frac{-1}{x^2} \end{align}
Equation 6.3.2.

Now let h be defined as $h(x)=x^{-n}$, n a positive integer. We can rewrite h as $h(x)=f(g(x))$, where $f(x)=x^n$ and $g(x)=x^{-1}$. By the chain rule, we have
\begin{align} f(x)&=x^n\\[1ex] g(x)&=x^{-1},x\neq 0\\[1ex] r_g(x)&=\frac{-1}{x^2}\\[1ex] h(x)&=f(g(x))\\[1ex] &=\left(g(x)\right)^n\\[1ex] r_h(x)&=n\left(g(x)^{n-1}\right)r_g(x)\\[1ex] &=n\left(\frac{1}{x}\right)^{n-1}\left(\frac{-1}{x^2}\right)\\[1ex] &=n\left(\frac{1}{x^{n-1}}\right)\left(\frac{-1}{x^2}\right)\\[1ex] &=\frac{-n}{x^{n+1}}\\[1ex] &=-nx^{-(n+1)}\\[1ex] &=-nx^{-n-1} \end{align}
Equation 6.3.3.

In Equation 6.3.3, we used the chain rule to generalize the power rule to $r_h(x)=-nx^{-n-1}$ when $h(x)=x^{-n}$ for positive integers $n$. Therefore, $r_h(x)=nx^{n-1}$ when $h(x)=x^n$ whether $n$ is negative or positive.

### Remarks on how to use the chain rule effectively

Deriving the rate of change function for a composite function is more than remembering the chain rule. The primary difficulty is recognizing that a function definition entails a composition of functions. There are two ways of thinking (WOT) that can help.

 WOT #1: Think of variables in the Library of ROC/Integral Pairs above as arguments instead of as variables. As variables they stand for numbers. As arguments they can stand for numbers, expressions, or functions. WOT #2: Look for structure in the definition of a function. Look for any expression that could define a function an expression that is raised to a power, an exponent that could be defined as a function, roots of an expression that could define a function, a product of expressions each of which could define functions.  An expression as simple as $x+1$ could define a function. The function h defined as $h(x)=(x+1)^2$ entails a composition. If could be written as $h(x)=g(f(x))$, where $f(x)=x+1$ and $g(u)=u^2$. Then $r_f(x)=1$ and $r_g(u)=2u^1$. By the chain rule $r_h(x)=r_g(f(x))r_f(x)$, or $r_h(x)=2(x+1)\cdot1$, which agrees with the result that we derived in Section 6.2.

Another question you will certainly have is, “When should I use the chain rule?”. The answer to this question is, “Always!!”. You can never go wrong by always using the chain rule. The worst that can happen is that your last rate of change function is 1, just like the example in WOT #2, above.

## Exercise Set 6.3.5a

1. This exercise is designed to let you practice WOT #1 above. Substitute an expression for x and then apply the chain rule. Use GC to check your answer. Keep in mind that you will not check your derived rate function against the approximate rate function for the function as given, but against the approximate rate function for your composite function.

For example: In (a), substitute $2x^2+1$ for x, getting $f(x) = (2x^2+1)^5$. Apply the chain rule to get the derived function $r_f(x)=5(2x^2+1)^4(4x)$. Check your answer by graphing your derived function and the graph of r defined as $r(x)=\frac{f(x+0.0001)-f(x)}{0.0001}$ (or using g and p as appropriate). If the two graphs do not agree, then it is likely that your derived function for the composite function is incorrect.

1. $f(x)=x^5$; $r_f(x)=5x^4$

2. $g(x)=2x^3-4x^{-2}+0.5x-1$; $r_g(x)=6x^2+8x+0.5$

3. $p(x)=(2x+1)^3$; $r_p(x)=3(2x+1)^2(2)$

2. This exercise is designed to let you practice WOT #2. Find the derived function r_ (the function name goes in the blank) for each function. Be alert for the possibility that you need to use the chain rule within using the chain rule in deriving r_. Use GC to check your answers.

1. $f(x)=(2(x+1)^2-2x-3)^3$

2. $g(x)=(x^4-(2x^2-2x)^{-2})^3+(x^3+2x^2-x-5)^2$

3. $h(x)=(((x^2+1)^3+2x)^2-3x^4)^3$

4. The function j is defined as $j(x)=f(g(h(k(x))))$ for functions f, g, h, and k that each have a rate of change function defined for all values of x. Use the chain rule to determine $r_j(x)$.

5. Relate Part d of this exercise to Parts a-c.

### Reversing the Chain Rule

Recall from Chapters 4 and 5 that if a function g has a rate of change at a moment for all values of x, then $dy=r_g(x)dx$ for all moments of x, where $r_g(x)$ is the rate of change of g at the moment x. We can use this relationship between y and x, together with the chain rule, to integrate rate of change functions that would look too messy to handle normally.

Let $g(x)=x^2+1$, and let $h(x)=3(x^2+1)^2\cdot2x$. Suppose that $h(x)=r_k(x)$ for some function k. If you look closely, you’ll see that  $2x=r_g(x)$, and therefore that $r_k(x)=3(g(x))^2 r_g(x)$. We know that $\int_a^x r_k(t)dt = k(x)-k(a)$. If we let $w=g(x)$, then $dw=r_g(x)dx$, and therefore:

 $k(x)-k(a)$ $=\int_a^x r_k(t)dt$ $=\int_a^x 3(g(t))^2 r_g(t)dt$ $=\int_{g(a)}^{g(x)} 3w^2 dw$ $= (g(x))^3-(g(a))^3$ $=(x^2+1)^3-(a^2+1)^3$

Equation 6.3.3a.

Therefore,  when $r_k(x)=3(x^2+1)^2 2x$, we get $k(x)=(x^2+1)^3$ by recognizing that $3(x^2+1)^2 2x$ is the rate of change function for a composite function, and then reversing the chain rule.

While there were several key moves in Equation 6.3.3a, the most critical one was to adjust the limits of the integral upon substituting w for $g(t)$. How might we understand the decision to make this adjustment? Like this: In $\int_a^x 3(t^2+1) 2tdt$, t varies from a to x. If we are to substitute w for $g(t)$ in this integral, how does w vary as t varies from a to x? Clearly, w varies from $g(a)$ to $g(x)$ as t varies from a to x. So, $$\int_a^x 3(g(t))^2 r_g(t)dt = \int_{g(a)}^{g(x)} 3w^2 dw.$$

In general, if you recognize that the rate of change function $r_k$ for an unknown function k has the structure $r_k(x)=r_f(g(x))r_g(x)$, and you recognize the functions f and g that have $r_f$ and $r_g$ as their rate of change functions, then:

$k(x)-k(a)$ $=\int_a^x r_k(t)dt$

 $=\int_{g(a)}^{g(x)} r_f(w)dw$, where $w=g(t)$

$=f(g(x))-f(g(a))$

Equation 6.3.3b.

Forthcoming.

## 6.3.6 Other Notations for Derived Functions

We have used the notation $r_f$ (called rate notation) consistently to represent a function f’s exact rate of change function. We did this continually to drive home the point that we are talking about rate of change functions—functions whose values give the rate of change at a moment of another function. Now that we are deriving rate of change functions, we’ll begin to use other notations, specifically $\frac{d}{dx}$ (called differential notation) and $f'$ (called prime notation). There is a fourth notational system, $D_f$ (called operator notation) that we’ll not use. Every notational system has advantages and disadvantages, and we will explain them for each.

Suppose that $f(x)=(x^2+1)^3$. Up to this point we represented f’s rate of change function using rate notation as $r_f$. By applying the derivations we have developed we see that $r_f(x)=3(x^2+1)^2(2x)$. Using differential notation we would write $$\frac{d}{dx}(x^2+1)^3=3(x^2+1)^2(2x).$$Using prime notation, we would write $$f'(x)=3(x^2+1)^2(2x).$$The advantage of rate notation is that we are reminded that the function we are representing is a rate of change function. The advantage of differential notation is that you see the function’s original defining formula in close proximity with the derived function’s formula. This makes it easier to remember rules, but it also runs the risk that you cease to think that you are finding a rate of change function. Prime notation is very similar to rate notation in that it reminds you that the derived function is a function and not the symbolic expression that defines it.

Differential notation is often used in the context of having defined a function without the use of function notation. If we write $y=(x^2+1)^3$, then we would write $\dfrac{dy}{dx}=3(x^2+1)^2(2x)$ and interpret it as giving us the momentary rate of change of the function defined by $(x^2+1)^3$ at every value of x.

Another advantage to differential notation is that it reminds us that we are talking about differentials in a function’s independent and dependent variables. The downside of differential notation is that it gives no way to represent the rate of change of the function at, say, $x=2$ other than to substitute 2 for x in $3(x^2+1)^2(2x)$. Moreover, we have not explicitly defined the original function nor its derived function. Rather, the focus is on having a formula to compute with.

An additional drawback of prime notation for us is that GC interprets the prime sign ($’$) as a command to draw a graph in a second graphing pane. To GC, prime notation has nothing to do with rate of change functions.

The chain rule is expressed differently in the different notational systems. Let f and g be functions and let $h(x)=g(f(x))$. To use differential notation we must also say to let $y=g(f(x))$ and $u=f(x)$. The chain rule would be expressed in the three notations as shown below. Keep in mind that all three statements say the same thing.

Let h be defined as $h(x) = g(f(x))$ for functions f and g, and assume that both f and g have a rate of change function that is defined for all values of x. Then the chain rule is expressed in the three notational systems as:

 Rate Notation Differential Notation Prime Notation $r_h(x)=r_g(f(x)r_f(x)$ $$\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}$$ $$h’(x)=g’(f(x))f’(x)$$

Do not read these statements literally (e.g., “dee y dee x equals dee y dee u times dee u dee x”). Instead, read them in terms of what they mean.
• Read the chain rule expressed in rate or prime notation as, “The rate of change of h with respect to x is the rate of change of g with respect to $f(x)$ times the rate of change of f with respect to x”.
• Read the chain rule expressed in differential notation as, “The rate of change of y with respect to x is the rate of change of y with respect to u times the rate of change of u with respect to x”.
We will use differential notation when emphasizing the relationship between a function’s defining formula and the defining formula of its rate of change function. We will use prime notation interchangeably with rate notation when our emphasis is on the fact that a derivative (derived function) is a rate of change function for an accumulation function.

## Exercise Set 6.3.6

1. Repeat the exercises in Exercise Set 6.3.2, but produce your work two times, first using differential notation and second using prime notation.

## 6.3.7 Rate Functions for Trigonometric Functions

### Sine and Cosine

Recall from Chapter 2 the meaning of θ and y in $y=\sin(θ)$: θ is the an angle measure in radians and y is a fraction of a circle’s radius. The animation in Figure 6.3.4 shows sin(θ) and cos(θ) varying in relation to θ. Focus on sin(θ) in relation to θ. Examine where sin(θ) varies faster and slower in relation to changes in the value of θ. Pause the animation and move the playhead manually to get a sense of how sin(θ) changes for equal increments of θ. Then do the same for cos(θ).

Figure 6.3.4. sin(θ) and cos(θ) vary in relation to the value of θ as it varies.

Reflection 6.3.3. Over what intervals in θ of length 0.5 radians does sin(θ) vary the most? The least? Why? Keep in mind that, for example, -0.75 is less than -0.5.

We can derive closed form representations for f and g defined as $f(x)=\sin(x)$ and $g(x)=\cos(x)$. In doing so, we will use two identities from trigonometry and two new insights into the sine and cosine functions.

The trigonometric identities are $\sin(u+v)=\sin(u)\cos(v)+\sin(v)\cos(u)$, which is used in the derivation of sine’s rate of change function in Equation 6.3.4, and $\cos(u+v)=\cos(u)\cos(v)-\sin(u)\sin(v)$, which is used in the derivation of cosine’s rate of change function in Equation 6.3.6.

Let $f(x) = \sin(x)$. We start the derivation of the closed form representation of f’s rate of change function by starting with its approximate rate of change function:

 $r(x))$ $=\dfrac{f(x+h)-f(x)}{h}$ $=\dfrac{\sin(x+h)-\sin(x)}{h}$ $=\dfrac{(\sin(x)\cos(h)+\sin(h)\cos(x))-\sin(x)}{h}$ $=\dfrac{\sin(x)(\cos(h)-1)+\sin(h)\cos(x)}{h}$ $=sin(x)\dfrac{(\cos(h)-1)}{h}+\dfrac{\sin(h)}{h}\cos(x)$

Equation 6.3.4.

In Equation 6.3.4 we came to line 5, where we need to understand the behavior of $\dfrac{\cos(h)-1}{h}$ and of $\dfrac{\sin(h)}{h}$ for values of h that are indistinguishable from 0 but not 0. We need to say that $h\neq0$ because neither expression is defined for $h=0$.

Look at the two frames in Figure 6.3.5. Figure 6.3.5a shows GC’s graph of $y=\dfrac{\sin(x)}{h}$. Figure 6.3.5b shows GC’s graph of $y=\dfrac{\cos(x)-1}{x}$. GC’s graph of $y=\dfrac{\sin(x)}{x}$ suggests that $\dfrac{\sin(x)}{x}$ is essentially 1 for values of x near 0. GC’s graph of $y=\dfrac{\cos(x)-1}{x}$ suggests that $\dfrac{\cos(x)-1}{x}$ is essentially 0 for values of x near 0.

 (a) (b)

Figure 6.3.5. $y=\dfrac{sin(x)}{x}$ and $y=\dfrac{1-cos(x)}{x}$. Examine behavior near $x=0$.

If we take as fact that $\dfrac{\sin(x)}{x} \doteq1$ for $h \doteq 0$ and that $\dfrac{cos(x)-1}{x} \doteq 0$ for $h \doteq 0$, then we can continue Equation 6.3.4 to conclude that
$$r_f(x)=\cos(x)\text{ when }f(x)=\sin(x)\text{ and that }\int_a^x \cos(t)dt = \sin(x) - \sin(a)$$
Equation 6.3.5.

From Equation 6.3.5 we can represent the rate of change function for sin(x) in three ways:

$r_f(x)=\cos(x)$ when $f(x)=\sin(x)$

$\dfrac{d}{dx}\sin(x)=\cos(x)$

$f'(x)=\cos(x)$ when $f(x)=\sin(x)$

We can determine a closed form representation of the rate of change function for $g(x)=\cos(x)$ as before: Start with an approximate rate of change function, derive a form that allows us to omit terms that involve h by letting h become so small that $h \doteq 0$.

 $r(x)$ $=\dfrac{g(x+h)-g(x)}{h}$ $=\dfrac{\cos(x+h)-\cos(x)}{h}$ $=\dfrac{(\cos(x)\cos(h)-\sin(x)\sin(h))-\cos(x)}{h}$ $=\dfrac{\cos(x)(\cos(h)-1)-\sin(x)\sin(h)}{h}$ $=\cos(x)\frac{\cos(h)-1}{h}-\sin(x)\frac{\sin(h)}{h}$, therefore $\doteq \cos(x) \cdot0 - \sin(x) \cdot 1$ when $h \doteq 0$ $\doteq -\sin(x)$ when $h \doteq 0$

Equation 6.3.6.

From the derivation in Equation 6.3.6 we can conclude that

$r_g(x)=-\sin(x)$ when $g(x)=\cos(x)$ and that $$\int_a^x -\sin(t)dt = \cos(x) - \cos(a)$$

Equation 6.3.7.

From Equation 6.3.6 we can represent the rate of change function for $\cos(x)$ in three ways:

$r_g(x)=-\sin(x)$ when $g(x)=\cos(x)$

$\frac{d}{dx}\cos(x) = -\sin(x)$

$g'(x)=-\sin(x)$ when $g(x) = \cos(x)$

Because the tangent function is the quotient of sine and cosine functions, we will derive the closed form rate of change function for the tangent function in the next section. It develops the rate of change function for product and quotent functions.

Version 3 of our Library of Rate of Change/Integral pairs is:

Library of ROC/Integral pairs, Version 3

 Accum Fn: $f(x)=$ Rate of Change: $r_f(x)=$ Accum Fn from Rate: ${\displaystyle \int_a^x r_f(t)dt}$ c 0 0 $cx^n$ $cnx^{n-1}$ $cx^n-ca^n$ $q(x)+p(x)$ $r_q(x)+r_p(x)$ $(q(x)+p(x))-(q(a)+p(a))$ $q(p(x))$ $r_q(p(x))r_p(x)$ $q(p(x))-q(p(a))$ $\sin(x)$ $\cos(x)$ $\sin(x)-\sin(a)$ $\cos(x)$ $-\sin(x)$ $\cos(x)-\cos(a)$

## Exercise Set 6.3.7

1. Find the exact rate of change function, $r_p(x)$, for each function $p(x)$ shown below. Check your derivation by graphing it along with the graph of p’s approximate rate function.

1. $p(x)=4(3x-8)^6+\sin(4\cos(x^5)+x^3-8)$

2. $p(x)=\cos(-(\sin(x))^2-\cos(x))$

3. $p(x)=\frac{1}{4}(\cos(x^2-4x+8))^8$

2. Suppose you have an exact rate of change function $r(x)=-\sin(x)+\frac{1}{2}\cos(x)+3x^3-8$. Find $\int_a^x r(t)dt$ in closed form.

3. Dale Earnhardt, Jr. is trying out a special device that records his speed throughout a NASCAR race.  The figure below shows a display of Dale’s speed, in $\mathrm {\frac{mi}{hr}}$, at each moment during an 80-second period of time.  Dale’s speed is modeled by the function s, which relates Dale’s speed, in $\mathrm {\frac{mi}{hr}}$, to the number of seconds that have elapsed.

1. What does the function s tell you about what Dale was doing during the first 8 seconds shown on the graph?

2. Explain a likely reason why $s(x)$ is sinusoidal from $x=8$ to $x=80$.

3. A lap at the Charlotte Motor Speedway is 1.5 miles.  Dale’s main competitor, Jeff Gordon, completed 2 laps in the first minute of this 80-second period.  Determine who went farther in the first minute.

4. Determine $r_s$, the rate of change function for s.  Check your derivation by graphing it along with the graph of s’s approximate rate function.
5. Use GC’s graph of $y=r_s(x)$ to state, approximately, the times at which Dale prepared to enter a curve in the track and when he started to come out of the turn, into a straightaway.

6. f)What does the graph of $y=r_s(x)$ tell you about s as a model of Dale’s speed during this time period?

4. A car has a piece of tape on a rear wheel so that we can record the number of radians that the wheel has swept after any amount of time has elapsed.  As the tire rotates, an angle of θ radians is swept out by the piece of tape from the 3 o’clock position. The angle’s measure increases at a rate of 3.4 $\mathrm{\frac{radians}{sec}}$. We started recording the tape’s height from horizontal at a moment when it was 0 cm.

5. Let h be the function such that $h(t)$ represents the tape’s vertical height, in cm, from the horizontal diameter, t seconds after recording began. So $h(0)=0$. The tire has a radius of 20.3 cm.

## 6.3.13 The Mean Value Theorem

Theoretical importance for justifying the work that we’ve done so far under the assumptions outlined in Section 6.3.12.