8 Chapter 8

2.3        Introductory Derivative Rules and Marginal Analysis

The derivative is a powerful tool, but as we saw in the previous sections, the process of finding the derivative using the limit definition is quite a tedious process. Fortunately (or unfortunately), one thing mathemati­ cians are good at is abstraction.

 

For instance, instead of working through the limit definition multiple times to find the value of the derivative at several x-values, we abstracted and worked through the process once to find a rule for the derivative function and then used this function to easily find the value of the derivative at several :r-values. Now, instead of finding the derivative of a function using the limit definition of the derivative, we will use formulas to quickly find derivatives of a variety of functions!

 

Although for this and the next two sections we will have to do some algebra that may seem tedious at times, an important thing to remember is that using the techniques we will learn to find the derivative is easier than using the limit definition of the derivative.

 

Learning Objectives:

In this section, you will learn how to use introductory derivative rules to calculate derivatives of functions and solve problems involving real-world applications. Upon completion you will be able to:

•   Calculate the derivative of a constant function.

•   Calculate the derivative of a power function.

•   Calculate the derivative of an exponential function (base e and base b).

•   Calculate the derivative of a logarithmic function (base e and base b).

•   Calculate the derivative of a function involving sums, differences, and constant multiples of constant, power, exponential, and/or logarithmic functions.

•   Calculate the slope of a tangent line using the Introductory Derivative Rules.

•   Find the equation of a tangent line using the Introductory Derivative Rules.

•   Graph a function and a line tangent to the graph of the function on the same axes.

•   Determine the :r-value(s) where the graph of a function has a horizontal tangent line using the Intro­ ductory Derivative Rules.

•   Determine the x-value(s) where the graph of a function has a tangent line of a given slope using the Introductory Derivative Rules.

•   Determine the :r-value(s) where a function has a specified (instantaneous) rate of change using the Introductory Derivative Rules.

•   Calculate the (instantaneous) rate of change of a function involving a real-world scenario, including cost, revenue, profit, and position, using the Introductory Derivative Rules.

•   Interpret the meaning of the derivative of a function involving a real-world scenario, including cost, revenue, and profit.

•   Calculate marginal cost, revenue, and profit using the Introductory Derivative Rules.

•   Interpret the meaning of marginal cost, revenue, and profit. – Estimate the cost, revenue, or profit of an item using marginal analysis and the Introductory Derivative Rules.

•   Compute the exact cost, revenue, or profit of an item.

•   Estimate the cost, revenue, or profit of a total number of items using marginal analysis and the Introductory Derivative Rules.

 

Introductory Derivative RulesFinding the derivative is, in a way, like analyzing a puzzle that has already been put together. We are given the whole picture (the function), and we need to identify the “pieces” of the function to take the derivative. Let’s start by examining the types of “pieces of the puzzle” we will see and how to find the derivative of each piece. Then, we will focus on how to combine the derivatives of the pieces to find the derivatives of more complicated functions.

 

Note: We will not be proving any of these rules rigorously, but we will explain them to some degree. We do want them to make sense after all!

Let’s start by examining the derivative of a linear function: f(x) = mx + b, where mis the slope and bis the y-intercept of the graph of the function. Keep in mind that the derivative, f'(:r), is the slope of the tangent line. When f is linear, the tangent line is the line itself! Because the slope of the line ism, we can conclude that f'(x) = m.

 

Now, let’s look at a constant function: f (:r) = k, where k is any real number. The graph of a constant function is a horizontal line, and the slope of a horizontal line is zero. So f'(x) = 0.

 

Next, let’s explore the relationship between g’ and f’ if g(x) = 4f(:r). This means the graph of g is four times as high as the graph of f. If we calculate the slopes of the secant lines of g and f, we will see that they follow this same relationship: The slopes of the secant lines of g are four times those off. This will

hold true for the tangent lines as well if we take the limit: The slopes of the tangent lines of g are four times those off! This is true for every tangent line at every point on the graph of g. This means the derivative of g is four times the derivative off. In other words, g'(:r) = 4f'(x). This gives us another rule:

 

Constants come along for the ride: lx (k • f (x)) = k ( lx (f (x))), where k is any real number. A complete list of the Introductory Derivative Rules, which includes the rules we have explored so far as well as the rules for combinations of functions, is given below:

 

Note: The Power Rule can be used for any power! This means it applies to negative exponents and rational exponents as well!

The Sum/Difference and Constant Multiple Rules combined with the Power Rule allow us to find the derivative of any polynomial and more!

Example 1 Find the derivative of each of the following functions.

(a)    f(x) = :r5

(b)   g(x)=1r518+10:r

(c)   J(t) = 30  –

(d) g(y) = 2.6y-4 – ln(9)

(e)   Q(x)=8.3ln(x)-xl.l+x15/4

(f)   P(x) = 14.223 (log44(:r) + 3x)

Solution:  (a) This function, f (:r) = :r5, is difficult to find the derivative of using the limit definition. It can be done, but the algebra gets very messy, very quickly.

Because this is a power function (the variable is in the base), we can use the Power Rule. We can think of the Power Rule as “bring the power down and reduce it by one”:

f(:r) = :i:5 ===;,

J'(x) = .5. x5-1

= .5a:4

(b)   First, notice g(a:) = 1r518 + 10:r consists of a sum of two functions, so we can apply the Sum Rule:

 

g'(a:) = –d    ( 1r”8 ) + –d(lOa:)

dx      dx

It is important to recognize that 1r5/8 is a number! It is approximately 2.0.5. We may be tempted

to use the Power Rule, but it falls under the Constant Rule because there is no x in the term.

Thus, lx (1r518) = 0. Now, the derivative becomes

 

g'(a:) = 0 + (lOa:)

d:r

d

= -(lOa:)

da:

Next, because we have a constant, 10, times a function, x, we can apply the Constant Multiple Rule and bring the constant to the front:

 

= 10 (d(x:r))

Due to the fact that x = a:1, we have a power function and can apply the Power Rule:

g'(:r) = 10 (1:J:1-1)

= 10 (a:0)

=10

 

To confirm our answer, notice that the original function is linear. Recall, as discussed previously, that the derivative of a line is just its slope. Because the slope of this line is 10, we can conclude that g'(a:) = 10.

(c)   First, notice f (t) = 30-   consists of a difference of two functions, so we can apply the Difference Rule:

f ,(t) = –d(3v5rt,_) – –d (

–et)

dt              dt     7

 

Next, notice that for each term we have a constant times a function. This is fairly easy to see for 30, but for ‘ it is easier to see once we rewrite it as tet. Now, we will apply the Constant Multiple Rule, which tells us to bring the constant to the front of each term:

f 1 (t) = d-(3vsrit) – d– ( 1–et)

dt             dt    7

dt               7    dt

= 3 (!£(vt)) – ! (!£ (et))

 

The Introductory Derivative Rules state that the derivative of et is et, but we do not have any

derivative rules for a root function. However, because we can rewrite a root function as a power function,0      = t115. Now, we can apply the Power Rule. Rewriting the root function and applying

the rules gives

All we have left to do is calculate the exponent of the first term by getting a common denominator:

1           1     5

–1=—

5           5     5

1-5

5

4

5

Therefore, J'(t) = Ir415– – Be careful when subtracting 1 from rational exponents. Take your time and be careful with the arithmetic.

(d)   For this function, g(y) = 2.6y-4 – ln(9), note that it does not matter what the independent variable is. We’ve seen x and t previously, but we may feel a little uncomfortable using y as the independent variable. We treat it the same as we would any other independent variable.

To find g'(y), first notice that g(y) consists of a difference of two functions. So we apply the Difference Rule:

 

For d  (ln(9) ), we may be tempted to say this equals ½, but ln(9) is a constant (like we saw with

1r5/8 in part b). Thus, iy (ln(9)) = 0. The derivative becomes

Because we have a constant times a power function, we apply the Constant Multiple Rule and bring the constant 2.6 to the front and then apply the Power Rule:

 

g'(y) = 2.6 ( d: (y-4))

= 2.6 (-4y-4-l)

= -10.4y-5

 

Be careful when finding the derivative of functions with negative exponents!  Remember that

-4 – 1 = -5, not -3.

 

(e)   The function Q(:r) = 8.3ln(x) – :i:1.1 + x15/4 has several terms subtracted and added. We apply the Difference Rule and Sum Rule to the appropriate terms. Immediately after that, we apply the Constant Multiple Rule to the first term:

QI (:r) = –d(8.3ln(:r)) –  –d    (:r1 •1 ) + –d ( :i.1:5)•

d:i.:                   d:i.:             dx

+

= 8.3 ( (ln(x)))-         (xl.1)         (x11-)

d:i.:                d:i.:            dx

 

Now, we apply the Logarithm Rule to the first term and the Power Rule to the second and third terms:

QI (x) = 8.3 (

-1) – l.lx1-•1 1 + 1–5x•15– 1

:i.:                         4

= -8.3

– l.lx0 •1

15    11

+

-x4

X                                4

(f)   The function P(:r) = 14.223 (log44(x) + 3x) consists of a constant times a function. Applying the Constant Multiple Rule gives

Notice inside the parentheses there is a sum, so we use the Sum Rule:

Applying the Logarithm Rule to the first term and the Exponential Rule to the second term gives

=                          +

1

P’ (:r)     14.223 (  l    ( )      3x ln(3))

:i.: n 44

Algebraic Manipulation

Notice that in part f of the previous example, we could have distributed the 14.223 first and then applied the Introductory Derivative Rules. We did not need to that, but there are functions in which algebraic manipulation is necessary before we can find the derivative using the Introductory Derivative Rules. In the coming sections, we will learn rules to use when functions are multiplied or divided. But for now, we will algebraically manipulate such functions and apply the Introductory Derivative Rules.

Example 2 Find the derivative of each of the following functions.

(a)   y =

(b)    f (x) = :i.:8 (x2- ;2)

(c)   g(x)= x”‘-2,:;3-llx

 

!

Solution: (a) We currently do not have a rule for division (that will come in the next section), but that does not mean we cannot find the derivative of y = using the Introductory Derivative Rules. But first, we have to do some algebra. Namely, we need to remember that dividing by :r is equivalent to multiplying by x-1:

3

y= –

:r

y = 3J:-1

 

Now, we have a constant times a power function, so we can apply the Constant Multiple Rule and Power Rule:

y’ = .!!:_ (3J:-1)

dJ:

=3 (d :

1

(x- ))

– )

= 3 (-lx-1 1

= -3x-2

 

(b)   Similar to part a, we do not have a rule for finding the derivative of a product of two functions (yet!), so we start by distributing x8:

 

J(x) = :r8 (:r2 – .!_)

x2

s    2         7:rs

=X       ·X      -­

x2

=Xl_O7:1;6

 

Using the Difference Rule, Constant Multiple Rule, and Power Rule gives

 

(c)

Because we do not have a rule for division (yet!), we work to rewrite the function, g(:r)

Due to the fact that there is only one term in the denominator, we begin by “separating” g(x)

into three fractions and simplifying each using laws of exponents:

x10 – 2×6 –  llx

g(x) =            J;6

x10     2:r6     llx

x6            x6            :r6

= x4 –  2 – 11:r-5

 

Using the Difference Rule, Constant Multiple Rule, Power Rule, and Constant Rule gives

 

g'(x) = .!!  (x4 – 2 – llx-5)

dx

d              d           d

= dx (x4) – d:r (2) – d:r (nr-5)

= .!! (:r4) – .!! (2) -11    (.!!   (x-5))

dJ:            d:r               dJ:

= 4:1:4-1 –  0 – 11 (-5:r-5-1)

= 4:r3 + 55:r-6

(d)   Again, because h(x) = (2:r – 5×2) (14×3 – 22) consists of a product of two functions and we do not have a rule for this type of multiplication (yet!), we must algebraically manipulate the function. We multiply the functions (using FOIL) to get a polynomial in standard form:

h(J:) = (2:r) (14×3) + (2x)(-22) + (–5×2) (14J:3) + (-.Sx2) (-22)

= 28×4 – 44J: – 70J:5 + 110×2

= – 70J:5 + 28×4 + ll0J:2 –  44x

Using the Sum/Difference Rule, Constant Multiple Rule, and Power Rule gives

h'(x) = –d

dJ:

 

(-70J:5 + 28×4 + 110×2 –  44x)

d            5           d         4           d           2           d

= dJ: (-70J: ) + dJ: (28x ) + dJ: (110x )-        dJ: (44J:)

 

= -70

(d!

(x5

)) + 28

(d!

(J.:4)) + 110

(d!

2

(J: ) )-    44

(d!

1

(J: ))

– )                           – ) –                      – )

= -70 (5J:5-1) + 28 (4×4 1  + 110 (2×2 1       44 (1J.:1 1

= -350x4 + 112×3 + 220x – 44

 

Example 3 Find the equation of the line tangent to the graph of g(x) = 10 – x2 at x = 2, and graph the function and the tangent line on the same axes.

Solution: Recall that to find the equation of the tangent line at x = 2, we need the line’s slope and the point on the line when J: = 2.

Remember that the slope of the tangent line at x = 2 is the derivative of the function at x = 2. With that in mind, we first need to find the derivative of g. Using the Difference Rule, Constant Rule, and Power Rule gives

g'(J:) = .!! (10 – x2)

d;r,

d             d

= d:r (10) – dJ: (x2)

= 0 – 2J:2-1

= -2x

 

To find the slope of the tangent line at x = 2, we calculate g'(2):

g'(2) = -2(2)

= -4

 

Next, we need to find the point on the tangent line when :r = 2. Remember that the tangent line touches the function at :r = 2, so we can substitute a;= 2 into the original function g(x) = 10 – :r2 to get the y-value of the point:

g(2) = 10 – (2)2

=6

This tells us that the tangent line passes through the point (2, 6). Thus, the equation of the tangent line is

y – 6 = -4(x – 2) y = -4:r + 8 + 6 y = -4x + 14

The graphs of g(.r) = 10- :r2 and the line tangent to the graph of g at .r = 2, y = -4x + 14, are shown in Figure 2.4.1.

 

Figure 2.3.1: Graphs of g(x) = 10 – :1:2 and the tangent line y = -4:1: + 14

 

 

Example 4 Better Purchase, a technology store, has a price-demand function given by p(a:) = 300(0.997)x, where p(a:) is the price, in dollars, of each mouse when x mice are sold. Find the rate of change of price with respect to demand when the demand is 800 mice, and interpret your answer.

Solution: The rate of change of price with respect to demand is given by the derivative function p’. We will use the Constant Multiple Rule and Exponential Rule to find p'(x):

 

p'(x) =d : (300(0.997?)

 

= 300 (d  ((0.997?))

= 300(0.997? ln(0.997)

 

Substituting 800 for :r to find p'(800) gives

p’ (800) = 300(0.997)800 ln(0.997)

::::, -$0.08 per mouse We write a sentence to interpret our answer:

“When 800 mice are demanded, price is decreasing at a rate of $0.08 per mouse.”

 

 

Example 5 The height of a ball thrown upward is given by s(t) = -16t2 + 128t feet, where tis time in seconds.

Find the velocity of the ball after 7 seconds.

Solution: Recall that velocity is the derivative of the position function. So we find s'(t), or v(t), using the Sum Rule, Constant Multiple Rule, and Power Rule:

dt

v(t) = s'(t) = d (-16t2 + 128t)

128 (: (t ))

= -16 (:t (t2)) +                1

= -16 (2t2-1) + 128 (lt1-1)

= -32t + 128

 

Now, we substitute t = 7 into the derivative:

v(7) = s'(7) = -32 • 7 + 128

= -96 feet per second

 

After 7 seconds, the ball is moving downward at a rate of 96 feet per second.

Note: The previous example may look familiar because we did it in Section 2.2, Example 7. In that example, we used the limit definition to find the derivative. Notice how much easier it is to use the Introductory Derivative Rules than the limit definition!

Example 6 Various values of J(x), g(:r), J'(:r), and g'(:r) are given in Table 2.8. Use the information in the table to answer each of the following.

 

:r	f(:r)	g(:r)	f'(x)	g'(x)
-4	-14	-92	19	128
0	0	12	-1	0
4	-2	4	3	16
8	28	172	31	176

 

Table 2.8: Various values of J(x), g(:r), J'(x), and g'(x)

(a)   If c(:r) = g(x) – J(:r), find c’ (4).

(b)   If r(x) = 3f(x) + 9x, find r'(0).

 

Solution:  (a) We start by using the Difference Rule because c(x) is a difference of two functions. This means c'(4) = g'(4) – f'(4). We find these function values in the table and conclude

c'(4) = g'(4) – J'(4)

= 16-3

= 13

(b) Using the Sum Rule, Constant Multiple Rule, and Exponential Rule, we have

Substituting :r = 0 and looking at the table to find f’ (0) = -1 yields

r'(0) = 3f'(0) + 9° ln(9)

= -3 + ln(9)

 

 

Example 7 The graphs of the functions f and g are shown in Figure 2.3.2. Use the graphs to answer each of the following.

 

Figure 2.3.2: Graphs of the functions f and g

(a)   If h(x) = f(x) + g(x), find h'(0).

(b)   If .i (:r) = 4:r2 – 7g(x), find j’ (-2).

Solution:  (a) Because h(:1:) is the sum of two functions, the Sum Rule gives h'(:1:) = f'(:1:) + g'(:1:).  Thus,

h'(0) = f'(O) + g'(0).

Recall that f'(0) is the slope of the line tangent to the graph off  at x = 0, and g'(0) is the slope of the line tangent to the graph of g at :r = 0.

 

Near :r = 0, the graph of f consists of a linear segment. Thus, the slope of the tangent line at

:r = 0 is the slope of this line segment, which is 2. Hence, f'(O) = 2.

Next, the graph of g has a horizontal tangent line at x = 0. Because the slope of a horizontal line is zero, we conclude that g'(0) = 0. Therefore,

h'(0) = J'(0) + g'(0)

=2+0

=2

(b) Using the Difference Rule, Constant Multiple Rule, and Power Rule gives

 

j'(:1:) = ..!!_ (4::r2 – 7g(:r))

d:r

= 4 (..!!_ (x2))  – 7 (..!!_ (g(:r)))

dx                    dx

= 4 (2×2- l)  –  7 (g’ ( X))

= 8:r – 7g’ (:r)

 

Substituting -2 for :r gives

 

j'(-2) = 8(-2)- 7g'(-2)

= -16 – 7g'(-2)

Near x = -2, the graph of g is linear. Thus, g'(-2) is the slope of this line segment, which is -1.

So g'(-2) = -1, and we have

j'(-2) = -16 – 7g'(-2)

=-16-7(-1)

= -9

 

Try  It 6

 

 

Example 8 Determine, algebraically, where y = 4:r113 + 15 is differentiable, and write your answer using interval notation.

Solution: To determine where the function is differentiable, we must find where its derivative, y’, is defined.

But, before we do that, we must first consider where the original function y is defined because its derivative will only (possibly) exist where y is defined. Looking at y, we see there are no domain restrictions, so the domain of y is (-oo, oo)

Next, we find y’. Using the Sum Rule, Constant Multiple Rule, Power Rule, and Constant Rule gives

 

To find where y’ does not exist, we have to consider the domain restrictions to find where y’ is undefined. At a glance it may appear that y’ has no “issues”, but remember that a negative exponent is another way to represent division:

 

Due to the fact that we cannot divide by zero,

2

3:r3 -/- 0

2

;r:s -/- 0

x-/-0

 

Because the domain of y is all real numbers and y’ will exist everywhere except at :r conclude that y is differentiable on (-oo, 0) U (0, oo).

Try  It 7

0, we can

 

 

 

 

Marginal AnalysisRemember that, at its heart, the derivative is the rate of change of a function. If that function is cost, revenue, or profit, we can determine how much the cost, revenue, or profit is changing per item when :r items are sold. This is referred to as marginal analysis.

Definition 2.7

 

Let’s look at some examples:

Example 9 Suppose that the profit, in dollars, obtained from the sale of x fish-fry feasts each week is given by P(;r:) = -0.03×2 + 8;1: – 50. Find the marginal profit when 64 fish-fry feasts are sold, and interpret your answer.

Solution: To find the marginal profit when 64 fish-fry feasts are sold, we need to calculate P'(64). We first find P'(x):

– )

– ) –

P'(;r:) = -0.03 (2;1:2 1 + 8 (1×1 1          0

 

 

Substituting 64 for :r gives

= -0.06:r + 8

 

P'(64) = -0.06(64) + 8

= 4.16

Recall that the units of the derivative function are the units associated with the output of the original function divided by the units associated with the input of the original function. Because P(x) is measured in dollars and x represents the number of fish-fry feasts sold, our units associated with 4.16 are dollars/feast. Thus, the marginal profit is $4.16 per feast.

 

We write a sentence to interpret our answer:

 

“When 64 fish-fry feasts are sold each week, profit is increasing at a rate of $4.16 per feast.”

 

Try It 8

 

Looking back at the previous example, notice our answer of $4.16 per feast is actually an estimate of the profit from selling the 65th fish-fry feast. This is because the derivative is a rate of change. It tells us the change in profit, in dollars, if one more item is made and sold. So if 64 feasts are currently sold and one more feast is sold (i.e., the 65th feast is sold), then profit increases by approximately $4.16. This is generalized below:

 

The profit obtained from making and selling the nth item can be approximated by P'(n – 1).

 

Now, let’s compare our estimate for the profit from selling the 65th feast ($4.16) to the exact profit from selling the 65th feast. To find the exact profit from selling the 65th feast, we only need to use the profit function, P(x) = -0.03×2 + 8x – 50. We find the profit from selling 65 feasts and subtract the profit from selling 64 feasts (this will leave us with the exact profit from selling the 65th feast):

P(65) – P(64) = (-0.03(65)2 + 8(65) – 50) – (-0.03(64)2 + 8(64) – 50)

= (343.25) – (339.12)

= $4.13

Note that our approximation of $4.16 is “close” to the exact profit of $4.13. The approximation has the benefit of being easier to compute (once we are comfortable finding derivatives). This is generalized below:

 

The exact profit obtained from making and selling the nth item is P(n) – P(n – 1).

 

Continuing on with our fish-fry fun, we can also use marginal analysis to approximate the profit from selling a total number of items. For example, if we want to estimate the profit from selling 65 feasts, we can find the profit from selling 64 feasts and then add the approximate profit from selling only the 65th feast (i.e., P'(64)). This will give us an approximation of the profit from selling all 65 feasts:

P(65);:;::P(64) +P'(64)

= (-0.03(64)2 + 8(64) – 50) + (-0.06(64) + 8)

= (339.12) + (4.16)

= $355.16

This is generalized below:

The profit obtained from making and selling n items, P(n), can be approximated by P(n – 1) + P'(n – 1).

 

Note: Because we are given the rule for P(x), using this approximation is more work than simply finding the exact profit from selling 65 feasts by calculatingP(65). You may be asking then, why do we need this approximation? \Nell, as we’ve seen in previous sections, we may not always be given a function. We may be given a graph or data points and be expected to find (or approximate) the profit. It is in those situations that this method becomes useful.

 

All of these ideas are similar for cost and revenue as well:

 

Estimating the Cost/Revenue/Profit of a Single Item Using Marginal Analysis

 

 

The  Exact Cost/Revenue/Profit of a Single Item

 

 

the Cost/Revenue/Profit of a Total Number of Items Using Marginal Anal­

ysis

 

The following examples demonstrate these ideas.

Example 10 A company that makes Barbara Dolls has a weekly cost function, in dollars, of C(x) = 10,000 + 90x –

0.05:r2, where x is the number of Barbara Dolls produced.

(a)   Estimate the cost of producing the 723rd Barbara Doll.

(b)   Find the exact cost of producing the 723rd Barbara Doll.

(c)   Approximate the cost of producing 723 Barbara Dolls

(d)   Find the exact cost of producing 723 Barbara Dolls.

Solution:  (a) First, note that we are looking for an estimate, not an exact answer. This means we should have a derivative in our calculation. Next, notice that we want to estimate the cost of producing a specific Barbara Doll (the 723rd) and not a group or total number of dolls. Looking at the formulas above, we see that to estimate the cost of the nth item, we find C’ (n – l). Here, we want

to estimate the cost of the 723rd item, so we calculate C'(723 – 1) = C'(722).

We must use the Introductory Derivative Rules to find C'(x), and then we can calculate C'(722):

– ) –                         – )

C'(x) = 0 + 90 (1:r1 1         0.05 (2×2 1

= 90- O.Lr

 

Substituting 722 for x gives

C'(722) = 90 – 0.1(722)

= $17.80 per Barbara Doll

 

Thus, the approximate cost of producing the 723rd Barbara Doll is $17.80.

 

Note: Remember, the derivative of the cost function gives us the change in cost if one more item is produced. The value of the derivative when 722 dolls are produced is our estimate for the cost of producing the 723rd doll because if the company currently makes 722 dolls and it makes one more doll, that would be the 723rd doll!

(b) We are asked to find the exact cost of producing a single Barbara Doll (specifically, the 723rd doll). Because this is an exact amount, our formula will not contain any derivatives. Using the

formula for finding the exact cost of producing the nth item, where n = 723, and the cost function

C(x) = 10,000 + 90:r – 0.05×2, we have

C(n) – C(n – 1) = 0(723) – 0(722)

= (10,000 + 90(723) – 0.05(723)2) – (10,000 + 90(722) – 0.05(722)2)

= (48,933.55) – (48,915.80)

= $17.75

 

Thus, the exact cost of producing the 723rd Barbara Doll is $17.75.

Note: This formula takes the total cost of making 723 Barbara Dolls and subtracts the total cost of making 722 Barbara Dolls. The result is the dollar amount it costs the company to make the 723rd doll. Note that this exact cost of $17.75 for the 723rd doll is relatively “close” to our estimate in part a, which we found to be $17.80.

(c) We need to approximate the total cost of producing 723 Barbara Dolls, which is a total number of Barbara Dolls, not just a single doll. Recall the formula for estimating the cost of producing n total items is given by

C(n) R:o C(n -1)   + C'(n – 1)

Substituting n = 723 and using the functions C(x) = 10, 000+90x-0.05×2 and C'(:r) = 90-0.1:r gives

0(723) R:o 0(722) + 0′(722)

= (10,000 + 90(722) – 0.05(722)2) + (90 – 0.1(722))

= (48, 915.80) + (17.80)

= $48, 933.60

Thus, the cost of producing 723 Barbara Dolls is approximately $48, 933.60.

Note: This estimate starts with the actual cost of producing 722 Barbara Dolls. Then, we add the value of the derivative when 722 dolls are produced. Why? The derivative gives us the change in cost if the company produces one more doll. So if it is currently making 722 dolls and makes one more doll, the company would be making 723 Barbara Dolls!

(d) To find the exact cost of producing 723 Barbara Dolls, we simply find the value of the cost function C(x) = 10,000 + 90:i: – 0.05×2 when x = 723:

0(723) = 10,000 + 90(723) – 0.05(723)2

= $48, 933.55

 

Note: This last part of the example is a reminder about what the cost function tells us. When substitut­ ing x = 723, the function outputs the cost of making 723 Barbara Dolls (no calculus required!).

 

Example 11 The company that made the Barbara Dolls in the previous example has a weekly revenue function given by R(:1:) = -0.02:r:2 + 45:1: dollars, where :r: is the number of Barbara Dolls sold.

(a)   Find the exact revenue from selling the 948th Barbara Doll.

(b)   Approximate the revenue from selling 1057 Barbara Dolls.

(c) Estimate the revenue from the sale of the 213th Barbara Doll.

 

Solution:  (a) The exact revenue from selling the 948th Barbara Doll is given by

R(948) – R(947) = (-0.02(948)2 + 4.5(948)) –  (-0.02(947)2 + 4.5(947))

= (24, 68.5.92) – (24,678.82)

= $7.10

 

(b) To approximate the revenue from selling 10.57 Barbara Dolls, we use the formula

R(l0.57)     R(l0.56) + R'(l0.56)

Recalling R(x) = -0.02×2 + 4.5x and using the Introductory Derivative Rules to find R'(x) gives

– )                         – )

R'(x) = -0.02 (2×2 1  + 4.5 (1×1 1

= –0.04x + 4.5

Therefore,

R(l0.57)     R(l0.56) + R'(l0.56)

= (-0.02(10.56)2 + 4.5(10.56)) + (-0.04(10.56) + 4.5)

= (2.5, 217.28) + (2.76)

= $2.5, 220.04

Thus, the approximate revenue from selling 10.57 Barbara Dolls is $2.5, 220.04.

(c) To estimate the revenue from the sale of the 213th Barbara Doll, we calculate

R’ (212) = -0.04(212) + 4.5

= $36.52 per Barbara Doll

 

Thus, the revenue from the sale of the 213th Barbara Doll is approximately $36.52.

 

Try  It 9

 

Try It Answers

1.     (a) ;l; = 12×2

(b)        ;l; = 170×9 + ex – 1.8

(c)

xln(3)

ddf x= -54x-10 – 2x ln(2) + 4 (-1-)

2.     (a) f'(:r) = -28×3 + 63:r2 + 22

(b)      g'(;1:) = 7tr  – ! – !;1:-3

12 ,

3• Y‘ – –  l;1·+43·,

 

4. -$0.43 per item; When 215 items are demanded, price is decreasing at a rate of $0.43 per item.

-5.  (a) 0

(b) 599/16

6.     (a) 18

(b) 2

7.   (-oo,0)U(0,oo)

8.   $33 per pair of shoes; vVhen 200 pairs of shoes are sold, revenue is increasing at a rate of $33 per pair of shoes.

 

9.  (a)	$25.04
(b)	$25.02
(c)	$17,500.02
(d)	$17,500

Exercises

 

Basic Skills Practice 

For Exercises 1 – 3, find f’ (x).

1.    f(x) = 4;r2

2.  f(;r) = -3:r3 + 1

For Exercises 4 – 6, find  .

4.   y = 6×2 – 4x + ex

5.   y = ln(x) – x7 + v’3

 

 

3.    f(x) = 5x-4 –  9:r + 12

 

 

 

 

6.   y = e; -17:r – ln(x)

For Exercises 7 – 9, find the slope of the line tangent to the graph of the function at the given ;1:-value.

7.   f (x:) = 9x:3 – ½:r + 1 at x: = -4

8.   g(;1:) = -¾:1:2 + ex – 6 at ;1: = 0

9.   h(x) = 2×5 + ln(x:) at x: = 1

For Exercises 10 – 14, find the equation of the line tangent to the graph of the function at the given x-value.

10.   h(x) = 2x-3 + fo: at x = 4

11.  g = 16:r2 –  9:i:3 + 11:r – 13 at :r = -2

12.   y = –l0x4 + 2ox-2 –  6 at at X = l

13.   f (x:) = ex + 4x5 –  5x:2 + 25 at x: = 0

14.   p(;1:) = 12×2 – x-6 + ln(x) + 2 at :r = 1

 

15.  Given g(t) = et +5, find the equation of the line tangent to the graph of g at t = 0, and graph both g and the tangent line on the same axes.

For Exercises 16 – 18, find the :i:-value(s) where the graph of the function has a horizontal tangent line.

16.    f(x) = 4;1:3 – 14;1:2 + 8;1: – 13

17.    f ( x) = 5×4 + 20×3 –  6

19.   Suppose that the profit, in dollars, obtained from the sale of ;1: barbeque dinners is given by P(x) =

-0.03×2 + 8x – 50. Find the rate of change of profit when 40 barbeque dinners are sold.

20.   The cost of producing :J’. calculators per week is given by C(:r) = 600 + lO0y’x dollars. Find the marginal cost when 250 calculators are produced each week.

21.   A flat iron manufacturer has a revenue function, in dollars, given by R(:1:) = 35;1: – 0.b:2, where ;1: is the number of flat irons sold.

(a)   Find the marginal revenue when 160 flat irons are sold.

(b)   Find the rate of change of revenue when 200 flat irons are sold.

22.   An arrow shot upward from the ground will have a height of s(t) = -16t2 + 128t feet after t seconds. Find the velocity of the arrow after 3 seconds.

23.   Various values of f(x),g(x),f'(x), and g'(x) are given in the table below. Use the information in the table to answer each of the following.

 

J’.	f (;1:)	g(;1:)	f’ (;1:)	g'(:r)
0	1	7	-4	2
1	-2	8	-2	0
2	-3	7	0	-2
3	-2	4	2	-4

 

(a)   If h(x) = f(x) + g(x), find h'(0).

(b)   If j(x) = 3g(:r), find j'(l).

(c)  If k(x) = -5g(x) +7f(x), find k'(3).

24.   A company that produces cameras has a cost function given by C(x) dollars when :r cameras are produced. Use the information in the table below to answer each of the following.

 

X	C(x)	C'(:r)
20	$1471.04	$22.94
21	$1494.43	$22.36
22	$1516.52	$21.82
23	$1538.08	$21.32
24	$1559.17	$20.85
25	$1580.15	$20.40

(a)   Find the rate of change of cost when 22 cameras are produced.

(b)   Find the marginal cost when 21 cameras are produced.

(c) Approximate the cost of producing the 23rd camera.

(d)        Find the exact cost of producing the 23rd camera.

(e)       Estimate the cost if 24 cameras are produced.

(f)       Find the exact cost if 24 cameras are produced.

2-5. A company that sells a popular lawn mower has a revenue function given by R(:-c) dollars when :r lawn mowers are sold. Use the information in the table below to answer each of the following.

 

:r	R(x)	R'(:r)
347	$116,633.64	$321.84
348	$116,955.84	$321.92
349	$117,277.96	$322.00
350	$117, 600.00	$322.08
351	$117, 921.96	$322.16
352	$118, 243.84	$322.24

(a)   Find the rate of change of revenue when 351 lawn mowers are sold.

(b)   Find the marginal revenue when 349 lawn mowers are sold.

(c) Estimate the revenue from selling the 350th lawn mower.

(d) Find the exact revenue from selling the 350th lawn mower.

(e) Approximate the revenue if 348 lawn mowers are sold.

(f) Find the exact revenue if 348 lawn mowers are sold.

26.     An appliance manufacturer that makes and sells high-end blenders has a profit function given byP(:r) dollars when x blenders are made and sold. Use the information in the table below to answer each of the following.

 

X	P(:r)	P'(:r)
1.50	$54,000.00	$385.00
151	$54. 384.70	$384.40
152	$54,768.80	$383.80
153	$55,152.30	$383.20
154	$.55, .535.20	$382.60
155	$-55, 917.50	$382.00

(a)   Find the rate of change of profit when 151 blenders are sold.

 

(b)   Find the marginal profit when 154 blenders are sold.

(c)   Approximate the profit from selling the 152nd blender.

(d)   Find the exact profit from selling the 152nd blender.

(e)   Estimate the profit if 153 blenders are sold.

(f)   Find the exact profit if 153 blenders are sold.

 

 

 

Intermediate Skills Practice 

For Exercises 27 – 33, find J'(x).

27.       f(x) = ,-Yx- 2×117 + lOex – 2ln(x)

28.     f(:r) = 6ln(x) -3:r2 + 25ex -5x318

29.     f(:r) = -½:r516 + 4:r2 + gx – 1

30.       J(x) = –7x3 + x213 – 5 · 4x – e3

For Exercises 34 – 37, fin*d     .

 

 

31.        f(x) = 2×3 – 7x + log4(:r) + v’19

 

33. J(x) = 0.03×0•2 + 183x-26/3 + 6. 10ox

 

 

 

36.   y = lOln(x) –          + x 5

37.

x-1/ 4

y = 8-½3 + 9 · 3ox –      TI

 

For Exercises 38 – 43, find the derivative of the function.

1.

x•

J(:r) = 3:r2 (x6 + 4:r3 – 5x + 2)                               4. y= 2×3-x10-21

2.

‘  –                 5t(j

g(x)=(x3+1)(x-6)                                               5. f(t) _ 3t5+2t-2-4t+s

3x-2

.‘}_,_ h(t) = (2c2 + 4t – 7) (3t + 1)                                6. h(:r)=  10×4-9x-3+15

 

For Exercises 44 – 47, find the slope of the line tangent to the graph of the function at the given x-value.

 

44. J(x) = 7:r6 – 4–½3 + 2log(:r) at x = 1 4-5. g(x) = 3ex – }s at x = 1

46. J(x) = (2x-3 + 4:r2) (:r-1 + 3:r) at x = -2

For Exercises 48 – 51 find the equation of the line tangent to the graph of the function at the given x-value.

 

48. f(:r) = 4ln(:r) – ;2 + 6       at x = 1

49. g(:r) = 2 · 10x + 3 at ;r = 0

-50.  f(:r) = (3:r4 –  x2 –  1) (6:c2 + :r) at :r = -2

3x_2

-51. h( x) = 2x-4-x6-5 at :r = 1

-52. Find the equation of the line tangent to the graph off (t) = 3 ln(t) + 2t2 – 1 at t = 1, and graph both

f and the tangent line on the same axes.

For Exercises 53 – 55, find the x-value(s) where the graph of the function has a horizontal tangent line.

5

53.  f(x) = 2ex – 5:r                        54. g(x) = (2×2 –  1) (4x + 1)           55. h(:r) = x8 – 2 ;: 7 +xG

 

For Exercises 56 – 58, find the x-value(s) where the graph of the function has the indicated slope.

56.   J(x) = 3:r – 8ln(x); slope is -10

 

57.   g(:r) = (x2 –  2) (x + 10); slope is 5

58.   h(x) = 3x5024x“; slope is 32

59.   Various values of f(:r),g(:r),f'(:r), and g'(:r) are given in the table below. Use the information in the table to answer each of the following.

 

X	f(x)	g(x)	f’ (:r)	g'(x)
-2	-28	-46	38	64
0	0	6	-2	0
2	-4	2	6	8
4	56	86	62	88

(a)   If h(:r) = -U(:1:) + g(:1:), find h'(-2).

(b)   If j(:1:) = ½x3 – 4g(:i:), find j'(4).

(c)   If k(:i:) = ,/2J(:1:) + 5:i:, find k'(0).

60.   The revenue function for a particular karaoke machine for kids is given by R(:1:) = 185:1: -0.2:i:2, where

R(x) is the revenue, in dollars, from selling x karaoke machines.

(a)   Find the rate of change ofrevenue when 225 karaoke machines are sold, and interpret your answer.

(b)   Find the marginal revenue when 500 karaoke machines are sold, and interpret your answer.

(c)   Approximate the revenue from selling the 145th karaoke machine.

(d) Find the exact revenue from selling the 145th karaoke machine.

(e)   Estimate the revenue if 300 karaoke machines are sold.

(f)   Find the exact revenue if 300 karaoke machines are sold.

61.   The profit function for a company that makes and sells designer shoes is given by P(x) = -0.3×2 +

475x – 10,500, whereP(:1:) is the profit, in dollars, when :1: pairs of shoes are made and sold.

(a)   Find the rate of change of profit when 975 pairs of shoes are made and sold, and interpret your answer.

(b)   Find the marginal profit when 625 pairs of shoes are made and sold, and interpret your answer.

(c)   Estimate the profit from making and selling the 700th pair of shoes.

(d)   Find the exact profit from making and selling the 700th pair of shoes.

(e)   Approximate the profit if 1532 pairs of shoes are made and sold.

(f)   Find the exact profit if 1532 pairs of shoes are made and sold.

62.   A company that produces dishwashers has a weekly cost function given by C(:r) = 3500 + 14:1: + 0.2:r2, where C(:1:) is the cost, in dollars, when x dishwashers are produced each week.

(a)   Find the rate of change of cost when 500 dishwashers are produced each week, and interpret your answer.

(b)   Find the marginal cost when 700 dishwashers are produced each week, and interpret your answer.

(c)   Approximate the cost of producing the 643rd dishwasher.

(d)   Find the exact cost of producing the 643rd dishwasher.

(e) Estimate the cost of producing 325 dishwashers.

(f) Find the exact cost of producing 325 dishwashers.

63.   A company’s total sales after t months is given by S(t) = 0.02t4 + 0.0:3t3 + 0.15t2 + 3t + 5 million dollars. Find the rate of change of sales after 8 months, and interpret your answer.

64.   La-Dreamer Dollhouses specializes in selling two-story dollhouses for kids. The company has deter­ mined its revenue function is R(x) = 210x – 0.8×2 dollars, where :1: is the number of dollhouses sold.

 

(a)   Find the marginal revenue function.

(b)   Find the marginal revenue when 75 dollhouses are sold, and interpret your answer.

6-5. A company has a cost function given by C(:r) = 2000 + lOx + 0.3×2 dollars, where xis the number of items produced. Find the company’s exact cost of producing the 50th item.

66.   The revenue, in dollars, when x digital clock radios are sold is given by R( x) = 10:r – 0.001:r2. Estimate the revenue from selling the 452nd clock.

67.   The profit function for a skateboard manufacturer is given by P(x) = 30x – 0.3:r2 –  250 dollars, where

:r is the number of skateboards manufactured and sold.

(a)   Estimate the profit from manufacturing and selling the 30th skateboard.

(b)   Find the exact profit from manufacturing and selling the 30th skateboard.

(c)   Estimate the profit from manufacturing and selling 30 skateboards.

(d)   Find the exact profit from manufacturing and selling 30 skateboards.

68.   A toaster manufacturer has a revenue function given by R(:r) = 42:r – O.Lr2 dollars, where :r is the number of toasters sold. Estimate the manufacturer’s revenue from selling 250 toasters.

69.   The position of an object moving along a horizontal line is given by s(t) = t3 – 3t2 – 5t + 6 feet, where

t is time in seconds. Find the velocity of the object after 2 seconds.

 

 

 

Mastery Practice 

./t:)

.      d (  {;’t2 – st-1/4 – frrr2

70.   Fmd –d                      ,r;.               •

t                  6vt

71.   Find J'(x) if f(:r) = (3:r114 –  4ft) (-½x3/4 + v?).

 

72• F. d   dy     “f  –       x2/3+7x”·•-5’7×3

Ill       –dX

1 y –

x -2/3

 

73.

.                                9

Given y = -6(0.43r +

10

ln(x), find y’.

-4e2t + 7t213et – 18et

74.   Given J(t) =                 t                          , find J'(t).

e

75.   Given q(:r) = -3:r:2 (12:1:3 –  9v’x) (8         – 7:r:6) + 77r2, find q'(:r:).

 

76.

3

Find :t ( (    2 –  t  – 10r7) (t + 9t5)).

 

77.   Given h(x) = 8Z(4-x)-6×2 , find

:r 2(

4-x )

ddxh.

.             (3:i:3 – 4:r-2) ( fa – 4:r5)            ,

78.   Given y =                x3                            , find y.

 

For Exercises 79 and 80, find the equation of the line tangent to the graph of the function at the given value.

 

79.   f(x) = yX (7x-3•5 + ¼vx) (::r6 –  7) at :r = 1

 

For Exercises 81 and 82, find the x-value(s) where the graph of the function has a horizontal tangent line.

X  –              9

81. p(.)  _  2 log” (x)-8x

83.   Find the value(s) oft where the function g(t) = 2t2 –  8ln(t) has an instantaneous rate of change of 4.

84.   The graphs of the functions f and g are shown below on the same axes. Use the graphs to answer each of the following.

(a)   If h(:r) = 0.24f(x) – g(:1:) + h2, find h'(-1).

(b)   If j(:1:) = g ) – ln(x) + v’7f(:1:), find j'(8).

(c)   If k(:1:) = 2f(x) – 3log5(x:), find k'(l).

(d)   If m(:r) = J(x) + (0.6)x + 4g(x) – e, find m'(2).

8-5. Find the values of the coefficients a, b, and c so that the parabola f (:r) = a:r2 + b:r + c passes through the point (1, 4) and the line tangent to the graph off  at the point (3, 14) is y = 9:r – 13.

86.   One Awesome Rainbow, a company that sells custom rainbow headbands, has a price-demand function given by p(:r) = 75(0.999?, where p(:r) is the price, in dollars, of each headband when x headbands are sold. Find the rate of change of price when 600 headbands are sold, and interpret your answer.

87.   The price-demand and cost functions for the production of cordless drills are given, respectively, by p(x:) = 143 – 0.03x: and C(x:) = 5,000 + 45x:, where :r is the number of cordless drills that are sold at a price of p(:r) dollars per drill and C(x) is the cost, in dollars, of producing :r cordless drills.

(a)   Find the marginal cost function.

(b)   Estimate the revenue from selling 2000 cordless drills.

(c)   Find the exact profit from producing and selling the 1050th cordless drill.

(d)   Find P'(lO00) and interpret your answer.

88.   It costs C(:r) = .jx dollars to produce :1:golf balls.

(a)   Find the marginal cost when 25 golf balls are produced, and interpret your answer.

(b)   Find the marginal cost when 100 golf balls are produced, and interpret your answer.

89.   A company that sells umbrellas has determined its price-demand function to be p(:r) = 25- .jx, where p(x) is the price, in dollars, of each umbrella when x umbrellas are sold.

(a)   Estimate the company’s revenue from selling the 50th umbrella.

2.3.  INTRODUCTORY DERIVATIVE RULES AND MARGINAL ANALYSIS                                    10-5

 

(b)   Find the company’s exact revenue from selling the 50th umbrella.

90.   A company has a price-demand function given by p(,r) = -2vx + 50, where p(x) is the price, in dollars, per item when :r items are sold. The company also has a cost function given by C(:r) = 90v:r +12x + 6,000 dollars when ;1: items are made. IfP(:1:) represents the company’s profit, in dollars, when ;1: items are made and sold, find P'(144). Interpret your answer.

91.   The position of a hummingbird flying along a straight line after t seconds is given by s(t) = 3t3 – 9t, where s(t) is measured in meters.

(a)   Find the velocity of the bird after 2 seconds.

(b)   When will the velocity of the bird be O meters per second?

92.   A company’s total sales, in millions of dollars, after t months is given by S (t) = 0.5t213 + 0.2yt + 6t + 4. Find the rate of change of sales after two years, and interpret your answer. Round to eight decimal places, if necessary.

93.   A hair dryer manufacturer has revenue and cost functions, both in dollars, given by R(:1:) = 38x-O.h2 and C(:r) = 5:r + 2000, respectively, where :r: is the number of hair dryers made and sold. Estimate the manufacturer’s profit from making and selling 175 hair dryers.

94.   A company has a revenue function given by R(x) = x(400 – 0.025x) dollars when x items are sold. Find the company’s marginal revenue when 200 items are sold, and interpret your answer.

9-5. A small business that sells customized candles can sell 100 candles each month when the price per candle is $25. For every dollar the price is raised, 10 fewer candles will be sold. The business has a cost function given by C(:1:) = 9:r + 1000 dollars, where x is the number of candles made. Assuming

the company’s price-demand function is linear,

(a)   estimate the profit from making and selling the 60th candle.

(b)   find the exact profit from making and selling the 60th candle.

(c) estimate the profit from making and selling 60 candles.

(d) find the exact profit from making and selling 60 candles.

 

 

 

Communication Practice 

96.     Explain why -{k (et) =I= tet-l_

97.     Explain why ix ( 5 =/- 5n4.

1r )

98.     If P(:r) is the weekly profit, in dollars, when :r items are sold each week, interpret P(lO0) + P’ (100) =

15,000.

d (3:r2 – 5x + VX) 6x – 5x ln(5) +l2x-1/2

99.     Does –                                =

dx             x4

4x3

? Why or why not?