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
accumulationrate 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,
$f(x)$ 
$=c$ 
$r(x)$ 
$=\dfrac{f(x+h)f(x)}{h}$ 

$= \dfrac{cc}{h}$ 

$= 0$ 
$r_f(x)$ 
$= 0$ 
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, $\int_a^x 0dt = f(x)f(a) = cc=0$.
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:
$f(x)$ 
$=4x^3$ 
$r(x)$ 
$=\dfrac{f(x+h)f(x)}{h}$ 

$=\dfrac{4(x+h)^34x^3}{h}$ 

$=\dfrac{4(x^3+3x^2h+3xh^2+h^3)4x^3}{h}$ 

$=\dfrac{4 \cdot 3x^2h+4 \cdot 3xh^2 + 4h^3}{h}$ 

$=4 \cdot 3x^2+4 \cdot 3xh + 4h^2$ 
$r_f(x)$ 
$ \doteq 4 \cdot 3x^2$ when $h \doteq 0$ 
Generalization:
$f(x)$

$=cx^n, c \neq 0, n \ge 0$ 
$r(x)$

$=\dfrac{f(x+h)f(x)}{h}$ 

$=\dfrac {c(x+h)^n  cx^n}{h}$


$= \dfrac{c(x^n + nx^{n1}h + n(n1)x^{n2}h^2 + ... + nxh^{n1} +
h^n)  cx^n}{h}$ 

$= \dfrac{cnx^{n1}h + cn(n1)x^{n2}h^2 + ... + cnxh^{n1} +
ch^n)}{h}$ 

$= cnx^{n1} + cn(n1)x^{n2}h + ... + cnxh^{n2} + ch^{n1}$ 
$r_f(x)$ 
$ \doteq cnx^{n1}$ 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 $y=4x^34a^3$
coincide. What does this tell you?
Check Generalization:
Let $f(x)=cx^n$. Use GC to see whether the graph of $y = cnx^{n1}$
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 $y=\int_a^x cnt^{n1}dt$ and
$y=c(x^na^n)$ 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

$cx^n$

$cnx^{n1}$

$cx^nca^n$

1) Let $g(x)=4$. What is $\int_a^x g(t)dt$?
Rewrite g as $g(x)=4 \cdot 1x^0$. Then $\int_a^x g(t)dt = \int_a^x 4
\cdot 1t^0dt = 4(x^1a^1)$.
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 $\int_a^x 4dt = 4(xa)$.
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 $\int_a^x 7t^3dt =
\int_a^x \frac{7}{4}(4t^3)dt=\frac{7}{4}(x^4a^4)$.
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,
$k(x)$

$=f(x)+g(x)$ 
$r(x)$

$=\dfrac{k(x+h)k(x)}{h}$ 

$=\dfrac{(f(x+h)+g(x+h))(f(x)+g(x))}{h}$


$=
\dfrac{(f(x+h)f(x))+(g(x+h)g(x))}{h}$ 

$= \dfrac{(f(xh)f(x))}{h} +
\dfrac{(g(xh)g(x))}{h}$ 
$r_k(x)$ 
$\doteq r_f(x) + r_g(x)$ 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

$cx^n$

$cnx^{n1}$

$cx^nca^n$ 
$q(x)+p(x)$ 
$r_q(x)+r_p(x)$ 
$(q(x)+p(x))(q(a)+p(a))$ 
Application
Let $h(x)=5x^23x+4$. 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 $k(x)=4$ 
$r_h(x)$ 
$=r_f(x) + r_g(x) + r_k(x)$ 

$= 10x + 3 + 0$ 

$= 10x3$ 
$\text{Let }h(x) = f(x)+g(x)+k(x) \text{ where }f(x)=5x^2, g(x)=3x^1\text{,
and }k(x)=4x^0.$
We can use the FTC to find $\int_a^x h(t)dt$ in closed form:
$$\begin{align}
\int_a^x h(t)dt &= \int_a^x f(t)dt + \int_a^x g(t)dt +\int_a^x
k(t)dt\\[1ex]
&= \int_a^x 5t^2dt + \int_a^x 3t^1dt +\int_a^x 4t^0dt\\[1ex]
&= \int_a^x \frac{5}{3} \cdot 3t^2dt + \int_a^x \frac{3}{2} \cdot
2t^1dt +\int_a^x 4t^0dt\\[1ex]
&= \frac{5}{3} \int_a^x 3t^2dt  \frac{3}{2}\int_a^x 2t^1dt +4\int_a^x
t^0dt\\[1ex]
&=\frac{5}{3}(x^3a^3)\frac{3}{2}(x^2a^2)+4(xa)\end{align}$$
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 $\int_a^x 3t^2dt=x^3a^3$.
Exercise Set 6.3.4
 Given that $f(x)=\dfrac{1}{2}x^32x+4$, 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 $h=0.001$.
 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 $y=r_h(x)$.
 $h(x)=5x^410x+2$
 $h(x)=\dfrac{5}{3}x^21$
 $h(x)=\dfrac{1}{4}x^6 + \dfrac{1}{3}x^3  x^4$
 Write $\int_a^x h(t)dt$ in closed form for each function h in
Exercise 2. Use GC to check your answers.
 In one kind of chemical reaction two separate substances combine to
form a third substance. The function g, defined as
$g(t)=18t3t^2$, gives the accumulated amount of the third substance t
minutes since the reaction began.
 What is $r_g$, g’s exact rate of change function, in closed
form?
 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?
 Describe in your own words what (b) tells you about the production
of the third substance at the 2minute and 4minute mark. (adapted
from Kline, p. 35)
 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.
 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?
 What restrictions should we place on our dependent and independent
variables? Why?
 What is the maximum height that the red bird will reach? Use
your rate of change function to justify this solution.
 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.
 Compare the blue bird’s maximum height to the red bird’s maximum
height.
 The binomial coefficients, denoted ${n \choose k}$ where $${n \choose
k} = \frac{n!}{(nk)!k!},$$ are the coefficients of the powers of x
when $(x+1)^n$ is expanded. That is,
$$\begin{align} (x+1)^n & = {n
\choose 0}x^0 \cdot 1^n + {n \choose 1}x^1 \cdot 1^{n1} + {n \choose
2}x^2 \cdot 1^{n2} + ... + {n \choose {n1}} x^{n1} \cdot 1^1 + {n
\choose n}x^n \cdot 1^0 \\ & = {n \choose 0}x^0 + {n \choose 1}x^1
+ {n \choose 2}x^2 + ... + {n \choose {n1}} x^{n1} + {n \choose
n}x^n \\ \end{align}$$
In Section 6.2 we found
that when $f(x)=(x+1)^n, r_f(x) = n(x+1)^{n1}.$ 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 $(x+1)^n$).
 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.)
 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 $\mathrm{\frac{ft}{sec}}$.
 What does $\int_0^5 18dt$ represent in this situation?
Evaluate this integral as part of your explanation.
 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 $r_f(x)=nx^{n1}$ 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.
$$\begin{align}
f(x)&=\frac{1}{x}\\[1ex]
r(x)&=\frac{f(x+h)f(x)}{h}\\[1ex]
&=\frac{\left(\dfrac{1}{x+h}\right)\left(\dfrac{1}{x}\right)}{h}\\[1ex]
&=\frac{\dfrac{x(x+h)}{x(x+h)}}{h}\\[1ex]
&=\frac{xxh}{hx(x+h)}\\[1ex]
&=\frac{h}{hx(x+h)}\\[1ex]
&=\frac{1}{x(x+h)}\\[1ex]
r_f(x)&\doteq \frac{1}{x^2}
\end{align}$$
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
$$\begin{align}
f(x)&=x^n\\[1ex]
g(x)&=x^{1},x\neq 0\\[1ex]
r_g(x)&=\frac{1}{x^2}\\[1ex]
h(x)&=f(g(x))\\[1ex]
&=\left(g(x)\right)^n\\[1ex]
r_h(x)&=n\left(g(x)^{n1}\right)r_g(x)\\[1ex]
&=n\left(\frac{1}{x}\right)^{n1}\left(\frac{1}{x^2}\right)\\[1ex]
&=n\left(\frac{1}{x^{n1}}\right)\left(\frac{1}{x^2}\right)\\[1ex]
&=\frac{n}{x^{n+1}}\\[1ex]
&=nx^{(n+1)}\\[1ex]
&=nx^{n1}
\end{align}$$
Equation 6.3.3.
In Equation 6.3.3, we used the chain rule to generalize the power rule to
$r_h(x)=nx^{n1}$ when $h(x)=x^{n}$ for positive integers $n$. Therefore,
$r_h(x)=nx^{n1}$ 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
 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.
 $f(x)=x^5$; $r_f(x)=5x^4$
 $g(x)=2x^34x^{2}+0.5x1$; $r_g(x)=6x^2+8x+0.5$
 $p(x)=(2x+1)^3$; $r_p(x)=3(2x+1)^2(2)$
 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.
 $f(x)=(2(x+1)^22x3)^3$
 $g(x)=(x^4(2x^22x)^{2})^3+(x^3+2x^2x5)^2$
 $h(x)=(((x^2+1)^3+2x)^23x^4)^3$
 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 $r_j(x)$.
 Relate Part d of this exercise to Parts ac.
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:
$k(x)k(a)$ 
$=\int_a^x r_k(t)dt$ 

$=\int_a^x 3(g(t))^2 r_g(t)dt$ 

$=\int_{g(a)}^{g(x)} 3w^2 dw$ 

$= (g(x))^3(g(a))^3$ 

$=(x^2+1)^3(a^2+1)^3$ 
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.
Exercise Set 6.3.5b
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
 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 $y=\dfrac{1cos(x)}{x}$.
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

$cx^n$

$cnx^{n1}$

$cx^nca^n$ 
$q(x)+p(x)$ 
$r_q(x)+r_p(x)$ 
$(q(x)+p(x))(q(a)+p(a))$ 
$q(p(x))$ 
$r_q(p(x))r_p(x)$ 
$q(p(x))q(p(a))$ 
$\sin(x)$ 
$\cos(x)$ 
$\sin(x)\sin(a)$ 
$\cos(x)$ 
$\sin(x)$ 
$\cos(x)\cos(a)$ 
Exercise Set 6.3.7
 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.
 $p(x)=4(3x8)^6+\sin(4\cos(x^5)+x^38)$
 $p(x)=\cos((\sin(x))^2\cos(x))$
 $p(x)=\frac{1}{4}(\cos(x^24x+8))^8$
 Suppose you have an exact rate of change function
$r(x)=\sin(x)+\frac{1}{2}\cos(x)+3x^38$. Find $\int_a^x r(t)dt$ in
closed form.
 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
80second 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.
 What does the function s tell you about what Dale was doing
during the first 8 seconds shown on the graph?
 Explain a likely reason why $s(x)$ is sinusoidal from $x=8$ to
$x=80$.
 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 80second period. Determine who went farther in the first
minute.
 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.
 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.
 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?
 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.
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.
 Determine a rate of change function $r_h$ whose values are the
tape’s rate of change of height from the horizontal diameter with
respect to the number of seconds that have elapsed. Check your rate of
change function by graphing it along with the graph of your h’s
approximate rate function.$
 At what rate is h increasing with respect to time at the
moment that $t=35.7$ sec?
6.3.8 Rate Functions for Product and Quotient Functions
In this section we will derive the general form of rate of change
functions for an accumulation function that is a product of two functions
that each have exact rate of change functions. Because a quotient function
h defined as $h(x)=\frac{q(x)}{p(x)}$, $p(x)\neq 0$, can be
expressed as $h(x)=q(x)(p(x))^{1}$, we will treat quotient functions as a
special case of the product of two functions.
Let u be a product of functions f and g that each
have a rate of change function at every moment. So $u(x)=g(x)f(x)$. We
know that in finding u’s exact rate of change function we will
form an approximate rate of change function that has
$g(x+h)f(x+h)g(x)f(x)$ in its numerator. But this is not very helpful in
its present form. Suppose we were to factor out $g(x+h)$. We could do this
if we had $g(x+h)f(x)$ as a term. Well, we can add $g(x+h)f(x)$ and then
compensate by subtracting $g(x+h)f(x)$, too, in order to keep the
numerator equivalent to $g(x+h)f(x+h)g(x)f(x)$. This is like adding zero
when we add and subtract the same term. The numerator in its new,
equivalent form is
which can be simplified to
which is far more useful as a numerator when we form u’s
approximate rate of change function because it represents both a change in
g and a change in f.
Equation 6.3.8 employs the trick of adding and subtracting the same term
in order to put the approximate rate of change function’s numerator in a
more useful form. It ends with the general result that
$r_u(x)=g(x)r_f(x)+r_g(x)f(x)$ when u is the product of two
functions g and f that each have exact rate of change
functions.
Equation 6.3.8.
Figure 6.3.6 shows a more intuitive derivation of the rate of change
function for a product of functions. It uses area as a model of
multiplication. The rectangle has side lengths of $g(x)$ and $f(x)$, so
the area function A defined as $A(x)=g(x)f(x)$ gives the
rectangle’s area for every value of x. The question is at what
rate the area increases when we increase x by dx.
Figure 6.3.6. Using differentials to derive the rate of change
function for a product of functions.
Figure 6.3.6 shows the value of x increasing by a tiny amount dx.
Since both g and h have exact rate of change functions,
they increase at the essentially constant rates $r_g(x)$ and $r_f(x)$
respectively for sufficiently small values of dx and therefore
their lengths increase by amounts that are essentially $r_g(x)dx$ and
$r_f(x)dx$, respectively. The rectangle’s area therefore increases by
$g(x)(r_f(x)dx)+(r_g(x)dx)f(x)$ plus $r_g(x)r_f(x)dx^2$, with
$r_g(x)r_f(x)dx^2$ being insubstantial given that dx is a tiny
increase in x and therefore that $dx^2$ is effectively 0.
Therefore $r_A(x)=g(x)r_f(x)+r_g(x)f(x)$.
We can express the product rule in three ways, using rate, differential,
and prime notation:
$u(x))$ 
$=g(x)f(x)$ 

$=g(x)r_f(x)+r_g(x)f(x)$ 
$\frac{d}{dx}u(x)$ 
$=g(x)(\frac{d}{dx}f(x))+(\frac{d}{dx}g(x))f(x)$ 
$u'(x)$ 
$=g(x)f'(x)+g'(x)f(x)$ 
Application of the Product Rule: The tangent function
The tangent function is defined as $$\tan(x)=\frac{\sin(x)}{\cos(x)},
\cos(x)\neq 0,$$ which can be rewritten as a product:
$$\tan(x)=(\sin(x))(\cos(x))^{1}.$$ We can derive the tangent function’s
exact rate of change function by applying the product rule, the power
rule, and the chain rule, and using the definition of the secant function
as $\sec(x)=\frac{1}{\cos(x)}$.
$\tan(x)$ 
$=\dfrac{\sin(x)}{\cos(x)}$,
$\cos(x)\neq 0$ 

$=(\sin x)(\cos x)^{1}$ 
$\dfrac{d}{dx}\tan(x)$ 
$=(\sin x \dfrac{d}{dx}(cos x)^{1}+(\dfrac{d}{dx}(\sin x))(\cos
x)^{1}$ 

$=((\sin x)(1(\cos x)^{2})(\sin x))+(\cos x)(\cos x)^{1}$ 

$=\dfrac{(\sin x)^2}{(\cos x)^2}+1$ 

$=\dfrac{(\sin(x))^2+(\cos(x))^2}{(\cos(x))^2}$ 

$=\dfrac{1}{\cos(x))^2}$ 

$=(\dfrac{1}{\cos(x)})^2$ 

$=(\sec(x))^2$ 
Equation 6.3.9.
Therefore, $r_{\tan}(x)=(\sec x)^2$.
We can express the derived rate of change function for tan(x) in three
ways using our three notation schemes:
$r_{\tan}(x)=(\sec x)^2$
$\dfrac{d}{dx}\tan(x)=(\sec x)^2$
$\tan'(x) = (\sec x)^2$
We can also add our new results to
our Library of Rate of Change/Integral Pairs:
Library of ROC/Integral pairs, Version
4
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

$cx^n$

$cnx^{n1}$

$cx^nca^n$ 
$q(x)+p(x)$ 
$r_q(x)+r_p(x)$ 
$(q(x)+p(x))(q(a)+p(a))$ 
$q(p(x))$ 
$r_q(p(x))r_p(x)$ 
$q(p(x))q(p(a))$ 
$\sin(x)$ 
$\cos(x)$ 
$\sin(x)\sin(a)$ 
$\cos(x)$ 
$\sin(x)$ 
$\cos(x)\cos(a)$ 
$\tan(x)$ 
$(\sec x)^2$ 
$\tan(x)\tan(a)$ 
$p(x)q(x)$ 
$p(x)r_q(x)+r_p(x)q(x)$ 
$p(x)q(x)p(a)q(a)$ 
Exercise Set 6.3.8
 Use the product rule to show that $r_g(x)=3$ when $g(x)=3x$.
 We know that $x^p x^q = x^ {p+q}$. When $k(x)=x^p x^q$ we could
simplify $x^p x^q$ to write k as $k(x)=x^{p+q}$ and use the
power rule to find that $r_k(x)=(p+q)x^{(p+q)1}$. Nevertheless,
deliberately use the product rule to show that
$r_k(x)=(p+q)x^{(p+q)1}$ when $k(x)=x^{p+q}$.
 Write $$\int_a^x 3t^2 \cos(t) + (6t)\sin(t)dt$$ in closed form.
 A soda company uses 139 $\mathrm{cm}^2$ of aluminum to produce a
cylindrical soda can. Suppose they want to know the best way to
make this can (what dimensions it needs for a radius and height) in
order to maximize the volume that they can fit into this cylindrical
can.
 Recall the formulas for surface area and volume of a cylinder.
 Define a function h that represents the cylindrical can’s
height as a function of x, the can’s radius (in cm).
(Keep in mind that we have 139 $\mathrm{cm}^2$ of aluminum to use.)
 Use part (b) to define a function v whose values are the
can’s volume in relation to its radius.
 Define the rate of change function $r_v$. Values of $r_v$ are the
exact rate of change of the can’s volume at any moment of its radius
as the radius varies from 0 to its maximum value.
 Determine the value of the radius that produces a can with the
largest possible volume. What is the can’s height when its
volume is maximum?
6.3.9 Rate Functions for Functions Defined Implicitly
In Chapter 3, Section 3.10, we
distinguished between functions defined explicitly and functions defined
implicitly. In the equation $x^2xy+y^2=3$ we can envision x being
a function of y or y being a function of x over
suitably restricted intervals of x and y. Suppose that we
envision x as a function of y. This supposition has the
same effect as rewriting $$x^2xy+y^2=3\text{ as}$$
$$(g(y))^2g(y)y+y^2=3,$$ where we now interpre "3" as the constant
function $f(y)=3$. We do not know the actual definition of g,
nor do we know immediately how y must be restricted so that g
is a function. But we can proceed nevertheless with the assumption that g
is defined as a function of y and that 3 is a function of y
without the definition of g and without knowing ahead of time
how y must be restricted.
Assume that g is an accumulation function and that y is
its independent variable. We can then apply what we know about derived
rate of change functions to the equation $(g(y))^2g(y)y+y^2=3$. Since the
left side and right side are equal, we can do the derivation given in
Equation 6.17. Equation 6.17 includes the following moves. We:
 presume that x can be defined as a function of y (line 2);
 substitute $g(y)$ for x (line 3);
 equate the derived functions of the left and right sides of the
equation (line 4);
 apply what we know about derived functions of a sum of functions
(line 5), and we apply the chain rule throughout (lines 6, 7);
 collect terms and solve for $\dfrac{d}{dy}g(y)$ (lines 8, 9);
 substitute x for $g(y)$ to get
$\dfrac{d}{dy}g(y)=\dfrac{2y}{2x1}$ (in differential notation)
or $r_g(y)=\dfrac{2y}{2x1}$ (in rate notation).
Equation 6.3.10.
It might seem strange that we end with a rate function for $g(y)$ that
involves both x and y. However, what our definition of
$r_g$ says is that values of $r_g(y)$ depend on values of both x
and y. While it might seem more accurate to express $r_g$ as a
function of both x and y, this would not be appropriate. To define
$r_g$ as a function of x and y would imply that both x
and y are independent variables for $r_g$. But this is not the
case. Values of x are not independent of values of y.
Instead, values of x are determined by values of y.
We can test the derivation in Equation
6.3.10 in this way: If the rate function for g is indeed
correct, then for any point $(x_0,y_0)$ on the graph of $x^2xy+y^2=3$,
the function passing through that point with a rate of change of
$r_g(y_0)$ and $x=x_0$ should be tangent to the graph of $x^2xy+y^2=3$.
Figure 6.3.7 demonstrates the accuracy of our test. The first pane of
Figure 6.3.7 shows that the point (1.58869,1.84651) is on the graph of
$x^2xy+y^2=3$. The second pane of Figure 6.3.7 shows that with
$x_0=1.58869$, $y_0=1.84651$, and m (our hypothesized rate of
change of x with respect to y when $x=x_0$ and $y=y_0$),
calculated according to Equation 6.3.7, GC’s graph of $x=m(yy_0)+x_0)$
is indeed tangent to the graph of $x^2xy+y^2=3$ at (1.58869,1.84651).
(a)

(b)

Figure 6.3.7. Testing Equation
6.3.10's derivation of $r_g(y)$.
The method exemplified in Equation 6.3.10, deriving the rate of change
function for a function defined implicitly, is often called implicit
differentiation. It is important to realize that implicit
differentiation (deriving the rate of change function for a function
defined implicitly) is a method. It is not a rule.
An Application of Implicitly Derived Rate
Functions: Rational exponents
As an immediate application of the method of implicitly derived rate
functions we will address the issue that, to this point, the power rule
applies only to functions that are to an integer power.
Suppose that $y=x^\frac{p}{q}$, where p and q are integers
and $q \neq 0$. If $y=x^\frac{p}{q}$, then it is also true that
$y^p=(x^\frac{p}{q})^q$, or $y^q=x^p$. Using the method of implicitly
defined rate functions, we have
Equation 6.3.11.
Therefore, $$\frac{dy}{dx}=(\frac{p}{q})x^{\frac{p}{q}1} \text{ when }
y=x^\frac{p}{q}.$$
We could also write this as
$$\frac{d}{dx}x^{\frac{p}{q}}=\frac{p}{q}x^{\frac{p}{q}1},$$
or as
$$r_f(x)=x^{\frac{p}{q}1} \text{ when } f(x)=x^\frac{p}{q}.$$
All of these statements assume that p and q are
integers, $q\neq0$.
In other words, the power rule works for rational exponents as well as
integers. Indeed, the power rule works for all real numbers, but the
argument for the power rule working for all real numbers demands an
understanding of real numbers that is beyond the scope of an introductory
calculus course.
Another Application of Implicitly Derived
Rate Functions: Inverse Trigonometric Functions
In this development of rate of change functions for inverse trigonometric
functions, we shall use “asin(x)”, which is short for “arcsine of
x”, to denote the inverse sine function instead of the oftenused notation
“$\sin^{1}(x)$”. This is for two reasons: (1) students often confuse the
negative exponent to mean $\frac{1}{\sin(x)}$, and (2) GC uses the
notation asin(x) to mean the inverse sine functionthe arc which
produces $\sin(x)$. We will similarly use “acos(x)” in place of
“$\cos^{1}(x)$” and “atan(x)” in place of “$\tan^{1}(x)"$.
We will use several facts and relationships in the derivation of the rate
of change function for asin(x), which we will denote in differential
notation as $\frac{d}{dx}\mathrm{asin}(x)$. These are:
 $y=\mathrm{asin}(x)$ implies that $\sin(y)=x$, $1 \le x \le 1$,
$\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$. Take note
that y is an arc length (in radians) and x is the
value sin(y).
 $(\sin(y))^2+(\cos(y))^2=1$, and therefore $\cos(y)= \pm
\sqrt{1(\sin(y))^2}$
 GC’s graph of $x=\cos(y)$ shows us that $\cos(y) \ge 0$ when
$\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$. So, $\cos(y)=
\sqrt{1(\sin(y))^2}$ because, as the first bullet states, the value
of y is indeed between $\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$.
The deriviation of $\dfrac{d}{dx}\mathrm{asin}(x)$ is given in Equation
6.3.12. We derive the rate of change function for $\mathrm{asin}(x)$
implicitly by deriving the rate of change function for sin(y),
where $y=\mathrm{asin}(x)$.
Equation 6.3.12.
Or, in rate notation, $r_f(x)=\frac{1}{\sqrt{1x^2}}$ when
$f(x)=\mathrm{asin}(x), 1<x<1$.
It is left as exercises for you to use the same method as above to derive
the rate of change functions for acos(x) and atan(x).
Exercise Set 6.3.9

Confirm Equation 6.3.10 by clicking on the graph of $x^2xy+y^2=3$
at several points, recording their coordinates, and modifying the GC
statements in Figure 6.3.7 accordingly.

In Equation 6.3.10 we assumed that x is a function of y.
Repeat Equation 6.3.10 with the assumption that y is a
function of x, then test your derivation in the same way as
in Figure 6.3.7.

Repeat the derivation in Equation 6.3.9 twice, first using rate
notation then using prime notation.

Assume that $y=g(x)$ in each of ac. Find the exact rate of change
function $r_g$ for each. Use any notational system that is
convenient. Check your derivation by graphing it along with the
graph of g’s approximate rate function.

$y=x^2 y^3 + y^4 x$

$x=x^2y^2,x>0$

$\sin^2(x)+\cos^2(y)=\cos(x+y)$

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

$f(x)=\tan(\sqrt{x})$
 $f(x)=\cos^2 (5x^4)$
 $f(x)=x \frac{2003x}{\pi}$
 $f(x)=(\sqrt[5]{x^4})(x2x^4)$
 $f(x)=\frac{5x^36x+9}{2x7}$
 $f(x)=\frac{3 \sqrt[5]{x^2}}{2x8}+3(x8)^{4}\sec^2(x)$
 For each of 5c and 5d, use your work to find $\int_a^x r_f(t)dt$ in
closed form. What do you notice?

A 10foot ladder leans against a wall. A mischievous monkey
kicks the ladder so that its bottom slides along the floor (see
animation below). The bottom of the ladder is initially 0 feet from
the wall. The bottom moves away from the wall at a rate of 1.3
feet per second.

Define a function h whose values give the ladder’s height
on the wall at each moment in time that the ladder falls.
 When does the ladder hit the floor?
 Derive the exact rate of change function, $r_h$, whose values give
the rate at which the height of the ladder is changing at each
moment in time. Check your derived function by graphing it along
with the graph of h’s approximate rate function.
 Graph $y=r_h(x)$ over an appropriate interval, where x
is the number of seconds that the ladder has fallen. Does the
ladder’s top ever fall faster than would a ball were it released
from a height of 10 feet, falling under the influence of gravity?
 A man having a height of 1.75 meters stood under a street light,
then walked away it at a rate of 1.7 meters per second. The
streetlight is 6 meters above the street. The streetlight casts a
shadow in front of him.
 Define a function s that relates the length of the man’s
shadow to the distance he has walked from the streetlight at any
particular moment in time since leaving the streetlight.
 How long is the man’s shadow after 5 seconds?
 Define a function $r_s$, in closed form, whose values give the
rate at which the man’s shadow lengthens with respect to the number
of seconds since he left the streetlight. Check your derivation by
graphing it along with the graph of s’s approximate rate
function.
 Define an integral based on $r_s$ that gives the length of the
man’s shadow x seconds after leaving the streetlight. What
is the closed form representation of that integral?
 Use the method of deriving a rate of change function implicitly to
show that $r_f(x)=\frac{1}{1+x^2}$ when $f(x)=\mathrm{atan}(x)$. Use
these facts: $y=\mathrm{atan}(x)$ means that $\tan(y)=x$;
$(\sec(y))^2=1+(\tan(y))^2$. How should the value of x be
restricted in $r_f(x)$?
Check your derivation by graphing it along
with the graph of $y=\frac{f(x+0.001)f(x)}{0.001}$.
 Use the method of deriving a rate of change function implicitly to
show that $r_f(x)=\frac{1}{1x^2}$ when $f(x)=\mathrm{acos}(x)$. Use
these facts: $y=\mathrm{acos}(x)$ means that $\cos(y)=x$;
$\mathrm{acos}(x)+\mathrm{asin}(x)=\frac{\pi}{2}$. How should the
value of x be restricted in $r_f(x)$?
Check your derivation
by graphing it along with the graph of
$y=\frac{f(x+0.001)f(x)}{0.001}$.
 The diagram below shows a kite 100 ft above the ground moves
horizontally away from you at a rate of 8 feet per second. The kite’s
height does not change, but the angle made by the ground and the kite
string changes as the kite moves away from you.
 Define a function that gives the rate of change of the string’s
angle with the ground at every moment in time. Check your definition
by graphing it along with the graph of your function’s approximate
rate function.
 At what rate is the angle between the string and the ground
changing when 150 feet of the string has been let out?
 The animation below shows the hands of a large clock rotating
rapidly. The hour hand is 1.5 meters long; the minute hand is 4 meters
long. Define a function whose values give the rate at which the
distance between the tips of the hour and minute hands changes with
respect to the measure of the angle between them.
6.3.10 Rate Functions for Exponential and Log Functions
In this section we will first review the ideas of exponential and
logarithmic functions and then derive rate of change functions for them
in closed form.
Exponential Functions
Paal Payasam is an Indian dish made of milk and rice. The Legend
of Paal Payasam illustrates the immensity of exponential growth.
The legend is that Lord Krishna played a game of chess against King
Rhada for this bet: If Krishna won, the King must give him 1 grain of
rice on the board’s first square, 2 grains on the second, 4 grains on
the third, and $2^{n1}$ grains on the $n^{th}$ square, up to $n=64$.
Krishna won the game, and King Rhada quickly understood the foolishness
of his bet. He foresaw having to put 4.29 billion grains of rice on the
$33^{rd}$ square and 9.22 billion billion grains on the $64_{th}$
squarea number of grains of rice that would cover all of India.
The expressions $x^b$ and $b^x$ have similar forms, but they have
entirely different meanings and, as illustrated by the Legend of Pall
Paysam, entirely different behaviors.
When we say $y=x^b$, we are saying that for any value of x, the
value of y contains x as a factor b times. If
$b=2$, then y contains 2 factors of x, or $y=x \cdot x$.
The value of x varies, but the number of factors does not vary.
When we say $y=b^x$, we are saying that
for any value of x, y contains b as a factor
x times. The number of factors varies but the value of each
factor does not vary.
$y=x^b$

$y=b^x$

y has x as a factor
b times
x varies, b is constant

y has b as a factor
x times
b is constant, x varies

Functions of the form $f(x)=x^b$ are
called power functions, as you already know. In a power
function, the exponent of x is constant. It does not vary.
Functions of the form $f(x)=b^x$ are called exponential
functions. In an exponential function, it is the exponent that
varies.
It would be convenient if exponential functions obeyed the power rule.
However, they do not. Figure 6.3.8 shows the power rule applied to
$f(x)=2^x$ (see GC’s graph of $r_1$) compared to f’s approximate
rate of change function (see GC’s graph of $r_2$). We trust that the
approximate rate function with $h=0.001$ will be reasonably close to the
exact rate of change function for f. GC’s graph of $r_1$, which
uses the power rule on f, is not even close to GC’s graph of
$r_2$, f’s approximate rate function. The power rule does not
apply to exponential functions.
Figure 6.3.8. Power rule applied to an exponential function does not
match the function's approximate...
Your understanding of the derivation of the general form of an
exponential function’s rate of change function will depend upon how well
you recall properties of exponents. As a reminder:
Two special cases are when $a=0$ and when $n = 0$ in $a^n$. To be
consistent with the fact that $\frac{a^m}{a^n}=a^{mn}$, we define $a^0$
to be 1 when $n=m$ and $a \neq 0$. We define $0^0$ to be 1 so that the
graph of $y=x^0$ is continuous at $x=0$. All values of $x^0$ are then 1.
The derivation of the rate of change function for $f(x)=b^x$ is given in
Equation 6.3.13. The derivation starts by defining the approximate rate
function r, and deriving the statement
$r(x)=b^x(\frac{b^h1}{h})$. That is, the rate of change function for
$f(x)=b^x$ is a product of $b^x$ and the expression $\frac{b^h1}{h}$.
We cannot let $h=0$ because $\frac{b^01}{0}=\frac{0}{0}$, which is
undefined. The question we must answer, then, is whether
$\frac{b^h1}{h}$ represents a number when $h \doteq 0$ (that is, the
value of h is indistinguishable from zero but not equal to
zero).
Equation 6.3.13.
Figure 6.3.9 investigates the behavior of $\frac{b^x1}{x}$ around $x=0$
for various values of b from 0 to 500.

For $b=0$, $\frac{b^x1}{x}$ has $x=0$ as a vertical asymptote.

For values of $b>0$, it appears that $\frac{b^x1}{x}$ is
always essentially equal to some number for $x \doteq 0$, although
it is a different number for each value of b.
The main point is that $\frac{b^h1}{h}$ is essentially equal to some
number when $h \doteq 0$ and $b>0$. But that number depends on the
value of b. For the meantime we’ll represent this number as
$c_b$; c to remind us that $\frac{b^h1}{h}$ is a number for $h
\doteq 0$, and “sub b” to remind us that this number depends on
the value of b. We will discover later what $c_b$ is.
Figure 6.3.9. Behavior of $\frac{b^x 1}{x}$ for values of x near 0
and for various values of b.
When $f(x)=b^x, b>0$, we have that $r_f(x)=c_b b^x$, where the value
of $c_b$ depends on the value of b. For some value of b,
$c_b$, or $\frac{b^h1}{h}$, is essentially equal to 1 when $h \doteq
0$. To this value of b we give the special symbol e, in
honor of Leonard Euler. Therefore $c_e=1$ by definition.
The importance of the value of b that makes $c_b=1$ is that the
rate function for this exponential function then becomes $r_f(x)=f(x)$
when $f(x)=e^x$. In differential notation, $\frac{d}{dx}e^x=e^x$.
While the value of e is irrational, its value is approximately
2.71828 (to 5 decimal places). The history of attempts to approximate
the value of e is fascinating. See Eli Maor’s excellent
account of this history.
Logarithmic Functions
A logarithmic function is the inverse of an exponential function, and
thus its definition depends on the base of the exponential function for
which it is inverse. In other words,
For $b>0$, $\log_b(x)=y$ such that
$b^y=x$
So $\log_3(81)=y$ such that $3^y=81$. Therefore $\log_3(81)=4$, because
$3^4=81$. Also, $\log_{\frac{1}{3}}(81)=4$ because
$\left(\dfrac{1}{3}\right)^{4}=81$.
We can use the inverse relationship between $\log_b$ and $b^x$ and the
method of implicitly derived rate functions to find the derived rate of
change function for f defined as $f(x)=\log_b(x), b>0$.
Equation 6.3.14 starts by stating the inverse relationship between
exponential and logarithmic functions and then uses the method of
implictly derived rate functions and the chain rule to end with the
derived rate function for $y=\log_b(x)$.
Equation 6.3.14.
In the case where $b=e$, $c_e=1$, so we have
$$\frac{d}{dx}\log_e(x)=\frac{1}{c_e x}\text{, or}$$
$$\frac{d}{dx}\log_e(x)=\frac{1}{x}.$$
Reflection 6.3.4. Why do we include the
restrictions $b>0$ and $x>0$ in the definition of a logarithm
function? Hint: Suppose that we allowed $b≤0$ or $b=1$? What would go
wrong? What about values of $x≤0$?
Because $e^x$ has the special property that $\dfrac{d}{dx}
e^x=e^x$ and because $\log_e(x)$ has the special property that
$\dfrac{d}{dx}\log_e(x)=\dfrac{1}{x}$, e is called the natural
base of the exponential and log functions. The natural log is
denoted “ln” (from Latin, logarithmus naturalis), so in
the future we will write $\log_e(x)$ as $\ln(x)$, and we will
call $\ln(x)$ the natural log of x. 
Equation 6.3.15 gives us the opportunity to determine the value of this
ubiquitous constant $c_b=(\dfrac{b^h1}{h})$ for $h \doteq 0$.
In determining $c_b$ we will use the simplicity of the derived
rate function for $f(x)=e^x$ together with the fact that
$b^{\log_b(x)}=x$ for any value of $b>0$ and for any value of
$x>0$.
Reflection 6.3.5. Check for yourself that
$b^{\log_b(x)}=x$. Use the fact that log and exponential functions are
inverse functions:
1) Let functions f and g be defined as $f(x)=b^x$ and $g(x)=\log_b(x)$.
2) Then $f(g(x))=x, x>0$ because f is the inverse of g.
3) But $f(g(x))=b^{\log_b(x)}$, so $b^{\log_b(x)}=x$.
Now we use the fact that $b^x=e^{\ln b^x}$ to determine the value of
$c_b$ in $\dfrac{d}{dx}b^x=c_b b^x$.
$\dfrac{d}{dx}
b^x$ 
$=c_b
b^x \text{ (already derived)}$ 
$b^x$ 
$=e^{\ln b^x}$ 
$\dfrac{d}{dx} b^x$ 
$=\dfrac {d}{dx} e^{x \ln b}$ 

$=e^{x \ln b} \dfrac{d}{dx} x \ln
b$ (Chain rule) 

$=e^{x \ln b} \ln b$ 

$= (\ln b)b^x$ 
Therefore, $c_b$ 
$=\ln b$, or 
$\dfrac{b^h1}{h}$ 
$=\ln b \text{ for } h \doteq 0$ 
Equation 6.3.15.
In summary, and in differential notation:
Equation 6.3.16.
The case of $r_f(x)=\dfrac{1}{x}$ when $f(x)=\ln(x)$ deserves more
consideration.
$\ln(x)$ is defined only when $x>0$. However, $\dfrac{1}{x}$ is
defined when $x>0$ and when $x<0$. The question
therefore becomes,
“What about $$\int_a^x \dfrac{1}{t}dt$$
when a and x are both negative?”
Figure 6.3.10 addresses this question. It shows that $\int_a^x
\frac{1}{t}dt$ is defined for $a<0$ and $x<0$ as well as for
$a>0$ and $x>0$.
Figure 6.3.10 also demonstrates that the graphs of $\int_a^x
\frac{1}{t}dt$ coincides with the graph of $y=\ln(x)\ln(a)$ when a
and x are both negative.
We can therefore say that $\int_a^x \frac{1}{t}dt = \ln(x)\ln(a)$
when a and x are both negative or both positive. We
cannot integrate $\frac{1}{t}$ over an interval that includes 0 because
$\frac{1}{t}$ is undefined for $t=0$, and, as we will see later,
$\int_0^c \frac{1}{t}dt$ is infinite for any value of c.
Figure 6.3.10. $\int_a^x \frac{1}{t} dt$ compared to
$\ln(x)\ln(a)$ for $a,x<0$ and $a,x>0$.
Exercise Set 6.3.10

Make an updated library of ROC/Integral pairs.

Show that $\log_3(x)$ can be expressed as
$\dfrac{\ln(x)}{\ln(3)}$.

Show that $\log_b(x)$ can be expressed as
$\dfrac{\ln(x)}{\ln(b)}$. Use this fact to graph $y=\log_7(x)$.
 Find the exact rate of change function for each of (a) and (b),
below. Check your derivations by graphing them along with the graph
of $y=\dfrac{m(x+0.001)m(x)}{0.001}$.
 $m(t)=7\ln(8t)+t^2\dfrac{1}{2}$
 $m(r)=\log_{10}(2+\cos(r))$
 A logistic
population model often is used in the biological and
ecological sciences to model the growth of animal populations.
This model takes into account the population’s initial size $P_0$,
its growth rate r per time unit, the environment’s carrying
capacity K (the maximum number of animals the environment
can sustain), and the value of m (the moment when the
expected population will be half its carrying capacity).
Values of the function P, defined as
$$P(t)=\frac{P_0 K}{P_0 +
(KP_0)e^{r(tm)}}$$
give the population size after t months have elapsed.
$P_0$ is the initial population size, K is the carrying
capacity, and r is the exponential growth rate for the
population. Let $m = 50$.
 Identify parameters and variables in the function definition.
 Create a slider in GC that will allow you to analyze carrying
capacities ranging from 2000 to 4000.
 Create another slider in GC that will allow you to analyze
initial populations ranging from 100 to 1000.
 Create a third slider in GC that allows you to analyze growth
rates from 0.002 to 0.05.
 Explain how each parameter affects population growth.
 We would like to find an exact rate of change function $r_P$ so
that we can analyze how the rate of change of the population with
respect to the number of days elapsed affects the population’s
growth rate. Derive $r_P$, then define it in GC. Check your
derivation with the approximate rate function for P.
 Given an initial population of 100 individuals, a carrying
capacity of 2500, and $r=.35$, use GC to evaluate the population
after 42 months from time $t=0$.
 For this same population, compute $r_P(42)$ and explain what
this value represents about the population.
 How might a game warden make use of the graph of P’s
rate of change function?
 Brenan monitors the rate at which water is moving in a stream
along the Colorado River during a rainstorm. He places his
measurement instrument in the river; it records how fast the river
is flowing in $\mathrm{\frac{m}{sec}}$.
From his map of the river bed, Brenan estimates the crosssectional
area of the river at the point where he takes his measurements. He
then enters this cross sectional area into his instrument’s radio
controller. With the river’s crosssectional area and the rivers
speed, the instrument can compute the flow rate, in cubic
meters per second. The measurement tool also records how many
hours have elapsed since being placed in the river. For more
information if you are interested on how these measurements are
taken by hydraulic scientists go
here.
Brenan returned after a rainstorm to pick up his equipment. He
went back to the lab where he created a mathematical model of the
flow rate—a function whose values give the water’s flow rate at each
moment in time during the recording. Here is Brenan’s model:
$f(x)=15(x^{14})e^{x^{15}}$, where values of f are a flow
rate x hours after recording began.
 Approximate the time the river’s flow rate changes from
increasing to decreasing.
 Approximate the time the river’s flow changes from increasing to
decreasing.
 At approximately what moment in time does the river’s flow stop
accelerating and start decelerating? Does the flow begin to
accelerate again at any point? If so, approximately when
does this acceleration occur?
 Brenan’s boss is worried that the local dam has gained too much
water in its reservoir. He needs to know how much water passed
that point in the river during the recorded time. How much
water flowed past the instrument during this period of time?
6.3.11 Rate functions for functions h in the form
$h(x)=f(x)^{g(x)}$
In this section we will derive rate of change functions for functions of
the form $h(x)=f(x)^{g(x)}$. Functions in this form are not power
functions, since the exponent is not constant, and they are not
exponential functions, since the base is not constant. Briggs, Cochran,
Gillet, and Schultz (2011) propose to call functions of the form
$h(x)=f(x)^{g(x)}$ tower functions. We will adopt Briggs et
al.’s proposal.
Suppose that h is defined as $h(x)=g(x)^{f(x)}$ and that f
and g have rate of change functions $r_f$ and $r_g$. We do not
have a method for deriving h’s rate of change function.
Just as in much of mathematics, our question is not so much about how to
do something, but how to represent what we want to act upon so that we
can use the methods that we have. We ask, “How can we represent the
function $h(x)=g(x)^{f(x)}$ in an equivalent form so that we can use
existing methods?”
We can rewrite the definition of h using the identities
$g(x)=e^{\ln(g(x))}$ and $\ln(x^y)=y\ln x$ to put the definition of h
in a form where we do have methods to derive its rate of change
function.
Before we do this, however, we must point out that $g(x)$ and
$e^{\ln(g(x))}$ are not always equivalent. Because $\ln(u)$ is defined
only for values of $u>0$, the expression $e^{\ln(g(x))}$ is defined
only for values of x such that $g(x)>0$.
Figure 6.3.11 illustrates the constraint that $g(x)$ must be greater
than 0. The first pane in Figure 6.3.11 shows GC’s graphs of $y=x^2$ and
$y=e^{\ln(x^2)}$ being identical. The second pane shows that GC’s graphs
of $y=\sin(3x)$ and $y=e^{\ln(\sin(3x))}$ are identical only for values
of x such that $\sin(3x)>0$.
(a)

(b)

Figure 6.3.11. Look between each pair of panes. The statement
$y=e^{\ln(g(x))}$ produces same graph as $y=f(x)$, but only for
values of x such that $g(x)>0$.
Suppose that $h(x)=g(x)^{f(x)}$, $g(x)>0$, and f and g
have rate of change functions $r_f$ and $g_f$. We will use two
identities in the derivation of $r_h$, namely:
$\ln(x^y)=y\ln x, x>0$
$x=e^{\ln(x)},x>0$
We also will use the chain rule several times, and the derived functions
for products and logs. The derivation of $r_h$ where $h(x)=g(x)^{f(x)}$,
$g(x)>0$ is given in Equation 6.3.16.
Equation 6.3.16.
In particular, if $g(x)=x$ and $f(x)=x$, $x>0$, then $h(x)=x^x$ and
$r_h(x)=x^x (1+\ln x)$.
Reflection 6.3.6. Download this
file. Then do the following.
 Reflect on how the GC commands in the file check whether the
derivation in Eq. 6.3.16 is valid.
 Modify the definitions of g and f to test whether
the derived rate function works for other definitions of g
and f.
 Pay particular attention to intervals over which $g(x)≤0$.
 Finally, examine the graph of $y=k(x)$ in comparison to the graph
of $y=r_k(x)$ to see whether the behavior of $r_k$ makes sense in
light of the behavior of k.
Reflection 6.3.7. Examine GC’s graph of $y=x^x$. Why is it
reasonable to adopt the convention that $0^0=1$ even though, logically,
$0^0$ is undefined?
Exercise Set 6.3.11
 In the figure below, the graphs of $y=\sin(x)$ and $y=\cos(x)$, $0
\le x \le 8$ are displayed in the left graphing pane; the graph of
$y=(\sin(x))^{\cos(x)}$, $0 \le x \le 8$, is shown in the right
graphing pane. Why does the graph of $y=(\sin(x))^{\cos(x)}$
increase so dramatically where it does? (Think about a really small
number being raised to a negative power.) Why does the graph have
gaps where it does?

Examine GC’s graph of $f(x)=(\cos(x))^x$. (Zoom out so that you
can see at least $20<x<20$.)
 Explain, in terms of values of $\cos(x)$ and values of x,
why the graph of f behaves as it does.
 Determine $r_f$, the exact rate of change function for f.
Explain the behavior of $y=r_f(x)$ in terms of the behavior of
$y=f(x)$.
 Derive the exact rate of change function $r_k$ in closed form for
each function k shown below. Check your answers using this
GC file. Study the behavior of $r_k$ to see that it reflects
the behavior of k.
 $k(x)=(2x)^{3x}$
 $k(t)=e^{\ln(5t^32t^2+3t4)}$
 $k(u)=(u^48)^{7u \cos(3u^24)}$
 $k(t)=(t^2+4)^m$
 Derive the exact rate of change function $r_f$ in closed form for
each function f shown below. Check your answers using this
GC file. Study the behavior of $r_f$ to see that it reflects
the behavior of f.
 $f(x)=10^{\log(3x7)}$
 $f(x)=\ln(e^{\sin(x)})$
 $f(x)=(3x4)^{e^x}$
 $f(x)=\left(\dfrac{1}{(x5)^2}\right)^{\sqrt{x3}}$
6.3.12 Big Assumptions We Made
Assumptions we made about functions that have a rate of change function
— continuity, no cusps. Show graphs of wildly behaving rates of change,
e.g. $\sin(x)+.01\sin(100x)$. Then graphs of $r(x)$ for a discontinuous
function; functions that are differentiable but whose rate of change
functions are not smooth; absolute value functions; Blancmange and
Weirsrass functions.
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.