9 Chapter 9

2.4  The Product and Quotient Rules

In the previous section, we introduced derivative rules which allowed us to find the derivative of some functions without going through the tedious process of using the limit definition of the derivative. One rule we introduced was the Sum Rule for derivatives, which states that the derivative of a sum is the sum of the derivatives. This rule might lead us to think there is a similar rule for finding the derivative of a product of two functions. Let’s see if this is indeed the case by looking at the function f below:

f (:r) = (3:r2 + 1) (7:r3 + 4J:)

Notice f(x) is written as a product of two functions. To find f'(x), can we just take the derivative of each factor and multiply the results? Recall from the previous section that

–d

dJ:

 

(3:r:2 + 1) = 6x

d

and- (7:r:

dx

 

3 + 4:r) = 21:r:2 + 4

Is it possible, then, that the derivative of f is simply

(6:r) (2Lr:2 + 4) = 126×3 + 24x?

To test this theory, recall from Section 2.3 that we are able to algebraically manipulate the original function so that we can apply the Introductory Derivative Rules to find the derivative:

j(J:) = (3:r:2 + 1) (7×3 + 4x)

= (3×2) (7×3) + (3×2) (4x) + (1) (7×3) + (1)(4x)

= 21×5 + 12×3 + 7×3 + 4J:

= 21J:5 + 19J:3 + 4J:

We see that f(:r) is now written as a sum of functions, so we can apply the rules from the previous section to find j'(J:):

j'(J:) = 21 (5×4) + 19 (:3:r2) + 4(1)

= 105×4 + 57×2 + 4

Notice this answer is not the same as our hypothesized answer above:

105Jc4 + 57J:2 + 4 =/- 126J:3 + 24x

Thus, we have convinced ourselves that we cannot just take the derivative of each factor and multiply the results!

 

The derivative of a product is NOT the product of the derivatives: d :(f(x) • g(x)) =/- j'(J:) • g'(x)

Even though finding the derivative of a product is not as simple as finding the product of the derivatives, luckily there is a rule for finding the derivative of a product (as well as a rule for finding the derivative of a quotient)!

 

Learning Objectives:

In this section, you will learn how to use the Product and Quotient 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 product of functions using the Product Rule.

 

•   Calculate the derivative of a quotient of functions using the Quotient Rule.

 

•   Calculate the derivative of more complicated functions by combining the Product, Quotient, and Introductory Derivative Rules.

 

•   Calculate the slope of a tangent line using the Product and Quotient Rules.

 

•   Find the equation of a tangent line using the Product and Quotient Rules.

 

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

 

• Determine the x-value(s) where the graph of a function has a horizontal tangent line using the Product and Quotient Rules.
• Determine the x-value(s) where the graph of a function has a tangent line of a given slope using the Product and Quotient Rules.

• Determine the x-value(s) where a function has a specified (instantaneous) rate of change using the Product and Quotient Rules.
• Calculate the (instantaneous) rate of change of a function involving a real-world scenario, including cost, revenue, profit, and position, using the Product and Quotient 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 Product and Quotient Rules.
• Interpret the meaning of marginal cost, revenue, and profit.
• Estimate the cost, revenue, or profit of an item using marginal analysis and the Product and Quotient 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 Product and Quotient Rules.
• Calculate average cost, revenue, and profit.
• Interpret the meaning of average cost, revenue, and profit.
• Calculate marginal average cost, revenue, and profit using the Product and Quotient Rules.
• Interpret the meaning of marginal average cost, revenue, and profit.

The Product Rule

As discussed previously, calculating the derivative of a product of two functions is a little more involved than calculating the derivative of a sum or difference of two functions. We will use the Product Rule to find the derivative of a product of two functions:

The Product Rule can be stated as “the derivative of a product is the first function times the derivative of the second function, plus the second function times the derivative of the first function.” Proving this rule is beyond the scope of this textbook; the important thing is knowing how to use the Product Rule.

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

(a)   f(:r) = (3:r2 + 1) (7:r3 + 4.1:)

(b)   g(x) = xex

(c)   h(t) = 4t3 –  y’tln(t)

Solution:  (a) Notice this is the function from the motivating example at the beginning of this section. Applying the Product Rule gives

 

J’ (x) = (3×2 + 1) (d : (7:r3 + 4:r)) + (7×3 + 4x) (d : (3:r2 + 1))

= (3:r2 + 1) (2lx2 + 4) + (7×3 + 4x) (6x)

In this section, we will find the derivative using the derivative rules and then stop. We tend to not worry about simplifying our answer unless we need to use the derivative to continue solving a problem and we must simplify it to do so. However, we will simplify our answer here to show that it is equivalent to the expression we found in the motivating example using the Introductory Derivative Rules:

.f'(:r) = (3:r2) (21×2) + (3:r2) (4) + (1) (21×2) + (1)(4) + (7:r3) (6x) + (4:r)(6:r)

= 63:r:4 + 12×2 + 2lx2 + 4 + 42:r4 + 24:r2

= lOfo:4 + 57:r2 + 4

This is the same answer we found in the motivating example!

(b)   Unlike in part a, we cannot perform any algebraic steps that would allow us to use the Introductory Derivative Rules to find the derivative of g(x) = :rex. We must use the Product Rule, where xis the first function and ex is the second function:

 

g’ (:1:) = :1: ( <}_ (ex)) + ex ( <}_.(:1:))

d:r                    d:r

= J.:ex + ex(l)

= xex + ex

(c)   For this function, h(t) = 4t3 –  y’tln(t), notice there is a difference of two functions. So we start by applying the Difference Rule from Section 2.3:

 

h'(t) = !!:_ (4t3) –  !!:  (vtln(t))

dt             dt

 

We can use the Power Rule along with the Constant Multiple Rule to find the derivative of 4t3. Also, recall that we should always rewrite radicals as power functions in order to apply derivative

rules. Thus, we rewrite0   as t112. This gives us

dt

h’ (t) = 12t2 –  d     (t½ ·ln(t))

 

Because there is a product in the second term, we must use the Product Rule. The first function is t112 and the second function is ln(t). This gives

 

–

h'(t) = 12t2

:t (t-½

ln(t))

–

= 12t2

 

–

= 12t2

( (t½) (:t(ln(t))) + (ln(t)) (! (t½)))

(t½ ( ) +ln(t) ( c½))

 

= 1. 2t2

1

– -t2 –

 

1 _!

2 ln(t)

-t

t      2

= 12t2 – c½ –  c½  ln(t)

2

 

Notice that when we applied the Product Rule, we had to write the entire result in parentheses, and then in the final step, we distributed the negative sign. We must always be careful with negative signs because a small slip will cause us to get the wrong derivative!

 

 

 

Example 2 Determine, algebraically, where y = ;r:½ log7(:r) 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, the only domain restriction is

;r: > 0 because of the log7(;r:) term (recall the argument of a logarithmic function must be positive). So

the domain of y is (0, oo).

Now, we find the x-values where y’ is defined, and assuming these x-values are in the domain of the function y, we will be able to state where the function is differentiable. Thus, we start by finding y’. To do so, we must use the Product Rule:

 

Because we must continue and find the domain of y’, we need to simplify the derivative (some):

 

Considering domain restrictions, we see that x4/5 -/- 0 because we cannot divide by zero. Solving this inequality leads us to conclude that :r-/- 0. Also, x > 0 because of the log7(x) term. To satisfy both of these restrictions, y’ must have a domain of (0, oo). Because this domain is a subset of the domain of the orginal function (in fact, they are the same!), we can conclude the function y is differentiable on (0, oo). Note that this means the function is differentiable at every point in its domain.

 

 

The Quotient Rule

We discovered previously that the derivative of a product is not the product of the derivatives, but what about quotients? Let’s explore this topic by looking at the function f below:

 

f(x) = -5

:r3

Notice that f(x) is written as a quotient, or ratio, of two functions. To find f'(x), can we just take the derivative of the numerator and the derivative of the denominator and divide the results?

We know that

 

d  (5) = 0    and     Is it possible, then, that the derivative of f is simply

= 07

3:r2

Recall from the previous section that we are able to rewrite the original function as

–

f(x) =  5

x3

 

= 5x-3

Applying the Introductory Derivative Rules to find the derivative gives

Notice this answer is not the same as our hypothesized answer:

 

Thus, we have convinced ourselves that we cannot just take the derivative of the numerator and the deriva­ tive of the denominator and divide the results.

 

Note: The derivative of a quotient is NOT the quotient of the derivatives: dd ( f((x))) =/= f'((:r)) .

X       g :r          g’ X

The question now becomes how do we find d  ( {1:i)? We can develop a formula using the Product Rule.

Suppose we let Q(:r) = {/;j, and we want to find the derivative Q'(:r).  If we multiply both sides of

Q(x) = {1:i by g(x), we get Q(x) · g(x) = f(x), or f(x) = Q(x) · g(:r). This is a product, so we can use the

Product Rule to find f'(:r):

Because our goal is to find a formula for Q'(x), we solve for Q'(x):

g(x) · Q'(x) = f'(:r) – Q(:r) · g'(:r)

Q

‘( )  _   J'(x)-Q(x)·g'(x)

X –                   g(x)

Next, because we want a formula that only involves J(;r) and g(x), we replace Q(x) with how it was originally

define,d

gJ((xx))•.

 

 

Q'(:1:=)

 

f'(:r) –  f(;) · g'(:r)

                   g( )

g(x)

Getting a common denominator and algebraically manipulating this expression to avoid having a complex fraction in our result gives

 

Q’ (:1.:=)

J'(x) _  f(x)-g'(x)

g(x)

g(x)

g(x) . J'(x) _  f(x)·g'(x)

g(x)                            g(x)

 

 

g(:r)

g(x)·J’ (x)        J(x)-g’ (x) g(x)                                 g(x)

 

 

g(:r)

g(:r) · f'(:r) – f(:r) · g'(:r)

 

 

g(x)

g(:r) · f'(x) – f(x) • g'(x)

g(:r)               (g(:r) )2

This is called the Quotient Rule.

 

The Quotient Rule can be verbalized as “the bottom function times the derivative of the top function, minus the top function times the derivative of the bottom function, all over the bottom function squared.” We will practice using the Quotient Rule in the next example.

Example 3 Find the derivative of each of the following functions.

(a)    f(x) = ;3

b)

(

y = x+1

x-7

c)

(

g(x)=    Iog9(x)

 

fi-4×2-12

 

Solution:  (a) Notice this is the quotient from the motivating example. Applying the Quotient Rule gives

 

J’ (x) = dd ( 53 )

X       X

(:r3) ( d  (5)) – (5) Ux (x3)) (x3)2

(:r3) (0) – (5) (3×2)

J:6

-15×2

—-;:e

15

a:4

 

This answer is equivalent to the answer we found in the motivating example, f’ (x) = -15x-4.

Note: It is always a good idea to rewrite a function like this when possible to avoid using the Quotient Rule because there is less room for error when using the Introductory Derivative Rules instead.

(b)   The function y =    consists of a quotient of two functions. But in this case, we cannot rewrite the function in order to use the Introductory Derivative Rules because there is more than one term in the denominator. So we use the Quotient Rule:

 

y=‘

(J;-      7) Ux (;r + 7)) – (:r + 7) ( d  (:r – 7)) (:r – 7)2

(x – 7)(1) – (x + 7)(1)

 

 

(:r – 7)2

x-7-x-7

(:r – 7)2

-14

(x – 7)2

 

(c)   The function g(a:) = )x ;: 12 also consists of a quotient of two functions, so we apply the Quotient Rule. Also, there is a radical, so we rewrite it as a power function ( ,Ji = :1:112):

 

(x½ -4×2 -12) Ux (log9(x))) -(log9(x)) (d  (x½ -4×2 -12))

g1 (:r:) =                                                             2

(:d – 4J:2 – 12)

9

(A -4:r2 -12) (x, ( ))-(log9(:r:)) (½:c½ -8:r:)

 

(:d – 4×2 – 12)2

 

Note: As stated previously, we will refrain from simplifying this derivative. This example asks us to find the derivative, and that is what we have done. If we need to use the derivative for some application, it might be necessary to simplify it to continue solving the problem. But for this example, this is the answer we are looking for.

In the previous examples, each function involved only one application of the Product Rule or Quotient Rule. Sometimes a function may involve a product and a quotient, or even two products or two quotients. We will practice finding the derivative of these more involved functions in the next example.

Example 4 Find the derivative of each of the following functions:

 

 

(a ) g

 

( t)

 

7tlllet

=

3t4-2t3+t2-1

(b)   H(t) = (1:a;it) (ln(t) + 3t)

(c)    f(t) = (2t8 + 13t2) (log3(t)) 14t

Solution:  (a) Observe that g(t) is a quotient of two functions, so we apply the Quotient Rule:

q’t     _ (3t4 -2t3 +t2 -1) (‘lt (7t10et)) – (7t10et) (‘lt (3t4 -2t3 +t2 -1)) “()-                 (3t4-2t3+t2-1)2

d

Notice that there are only two parts that involve more work. We must find the derivative of 7t10et and the derivative of 3t4 – 2t3 + t2 – 1. The derivative of the latter just involves the Introductory Derivative Rules:

–  (3t4

dt

–  2t3

+ t2

– 1) = 12t3

–  6t2

+ 2t

Because 7t10et is a product of two functions, we must use the Product Rule to find its derivative:

:t  (7t10 el) = 7t10 (! (el)) + et (! (7t10))

= 7t10 et + et (70t9)

= 7t10 et + 70t9et

 

Substituting both of these into g'(t) gives

 

(b)   We start with the Product Rule because, at the heart of it, H(t) = u:a;it) (ln(t) + :3t) is a product of two functions:

 

2                                                                                                      2

H'(t) = ( t –      2.2 ) ( d-::,(ln(t) + 3t)) + (ln(t) + 3t) ( d-::, ( t –      22 ) )

4t3 + ’13t      at                                           at    4t3 + 8t

 

d                      1

Again, notice there are only two parts that involve more work. We must find the derivative of ln(t) + 3t and the derivative of J:a;it’ The derivative of ln(t) + 3t involves the Introductory Derivative Rules:

-::,(ln(t) + 3t) = – + 3

at                      t

Because 1:3;it is a quotient of two functions, we use the Quotient Rule to find its derivative:

–

–

!!_(  t2 –            22)=     (4t3 + 8t) ({ft (t2                 22)) – (t2               22) (ft (4t3 + 8t))

dt    4t3 + 8t                                      (4t3 + 8t)2

(4t3 + 8t) (2t) – (t2 –              22) (12t2 + 8) (4t3 + 8t)2

Substituting both of these into H'(t) gives

‘()      ( t2

 

22)(1     )      (     ()

3    8t) (2t-)    (t2                         2

H t =       –

– + 3  + ln t + 3t) ( (4t +

–      22)(12t

+ 8))

4t3 + 8t       t                                                    (4t3 + 8t)2

(c)

3

Notice f(t) = (2t8 + 13t2) (log3(t)) 14t is the product of three functions. We will start by applying the Product Rule to the two functions (2t8 + 13t2) log (t) and 14t:

Again, there are only two parts that involve more work.  We must find the derivative of 14t

and the derivative of (2t8 + 13t2) log3(t). Applying the Introductory Derivative Rules to find the derivative of 14t gives

!!_ (14t) = 14t ln(14)

dt

To find the derivative of (2t8 + 13t2) log3(t), we need to use the Product Rule a second time:

:t ((2t8 + 13t2) log3(t)) = (2t8 + 13t2) (!(log3(t))) + (log3(t)) (! (2t8 + 13t2))

 

3

= (2t8+13t2)  (tl:( )) +(log3(t))(16t7 +26t)

Substituting both of these into f’ (t) yields

3

J’(t) = ((2t8 + 13t2) log3(t)) (14t ln(14)) +(14t) ( (2t8 + 13t2) ( tl:( )) + (log3(t)) (16t7 + 26t))

 

Note: Note the difference when starting parts a and b. In part a, we started with the Quotient Rule because the most “outside” part of the function (or the last thing we would do when computing it) is division, whereas in part b, the most “outside” part of the function is multiplication. Thus, we started with the Product Rule.

 

Note: For some, the solutions in this section might seem a little overwhelming. It is highly recommended that when you approach these problems you take one step at a time. If you try to do it all at once, there is a greater chance of making a mistake. Whether you are applying the Product Rule or the Quotient Rule, you eventually are just applying the Introductory Derivative Rules from the previous section, which are much easier to compute.

 

Example 5 Find the equation of the line tangent to the graph of J(;r) = and the tangent line on the same axes.

30:rln(x) at x = 1, and graph the function

3x2 + 12

Solution: Remember, to find the equation of a tangent line (or any line), we need its slope and a point on the line. Thus, our first order of business is to find the derivative off  because we know that f'(l) will give us the slope of the tangent line at ;1: = l. f (;1:) is a quotient of two functions, so we apply the Quotient Rule:

, :r _ (3;1:2 + 12) U;,(30:i:ln(:r))) – (30;1:ln(:i:)) U{ (3:r2 + 12))

f (.)-                                   (3×2+ 12)2

 

We know that d  (3×2 + 12) = 6x, but because 30x ln(x) is a product of two functions, we must apply the Product Rule to find its derivative:

 

d :(30xln(x)) = 30x (d!(ln(x))) + (ln(x)) (d /30x))

 

= 30;1: ( )      + ln(;1:)(30)

 

= 30:r (  )  + 30 ln(:r)

= 30 + 30 ln(:r)

 

Substituting both of these into f'(x) gives

,          (3×2 + 12) (30 + 30ln(x)) – (30xln(x))(6x)

f   (:r)=                           (3:r2+ 12)2

 

Now, we substitute 1 for :r to get the slope of the tangent line at :r = 1:

,          (3(1)2 + 12) (30 + 30ln(l)) – (30(1) ln(l))(6(1))

f   (1)=                             (3(1)2+ 12)2

(15)(30 + 0) – 0

(15)2

=2

 

Next, we need to find a point on the line. We know x = 1, and we can find the corresponding y-value

3:r + 12

by                 the                           f(x) =                at :r = 1:

.              . .             .                30x ln(x) evaluatmg       ongmal funct10n         . 2

 

f(l)= :30(1) ln(l)

3(1)2 + 12

0

15

=0

 

Thus, the tangent line passes through the point (1, 0) and has a slope of 2. Using the point-slope form of the equation of a line gives

y-0=2(x-1)

y = 2x – 2

 

The graphs of f(x) = 30:r ln(:r) and y = 2x – 2 are shown in Figure 2.4.l.

3;1:2 + 12

 

 

2

Figure 2.4.1: Graphs of f(x) = 3i: i ) and the line tangent to the graph off  at :r = 1

Example 6 Table 2.9 shows values of f(:r), g(x), f'(x), and g'(x) for certain values of x. Use the information in the table to answer each of the following.

 

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

 

Table 2.9: Various values of J(x), g(:r), J'(x), and g'(x)

(a)

g X

If T(x) = .f((x)), find T'(8).

 

Solution:  (a) T(:r) is a quotient of two functions, so we will start by applying the Quotient Rule to find T'(:r):

Substituting x = 8 and then finding the appropriate function values in the table gives

T'(8)=  g(8) • f'(8) – f(8) • g'(8)

(g(8))2

(172) (31) – (28)(176)

(172)2

404        101

29584     7396

(b)   To find Z'(J.:), where Z(x) = J.:2 • g(x) – log3(x), we start with the Difference Rule:

 

To find the derivative of the first term of the difference, we must use the Product Rule. We will also use the Introductory Derivative Rules when finding both derivatives:

Z'(:r) = –d

dx

 

(:r2 •

d

g(x)) – –

dx

 

(log3(x))

= x2 • g'(:r) +

d

g(:r) · –    (x

2       d

) –  –

(log3(x))

dx            dx

= x2 • g'(:r) +g(:r) · (2:r) – -­ 1

x ln(3)

Substituting x = 4 into Z'(:r) = :r2 • g'(x) +g(:r) • (2x) – x!J(3) andthen finding the appropriate function values in the table gives

Z /(4) = (4)2 • g/(4) +g(4) • (2 • 4) –     l1 ( )

4 n 3

1

= 16 · 16 + 4 · 8 – —

4 ln(3)

288        1

=           –     4 ln(3)

 

Example 7 The graphs of the functions f and g are shown in Figure 2.4.2. Use the graphs to answer each of the following.

 

Figure 2.4.2: Graphs of the functions f and g

(a)        If h(i:) = f(i:) · g(i:), find h'(O).

(b)        If j(x) = g x)c–;r, find j'(3).

 

Solution:  (a) Using the Product Rule to find h'(x) gives

h'(i:) = J(x) • g'(x) + g(x) • J'(i:)

 

Substituting x = 0 we get

h’ (0) = f (0) • g’ (0) + g(0) • J’ (0)

Looking at Figure 2.4.2, we see f (0) = 5 and g(0) = 1. Recall that these are the y-values of the points on the graphs corresponding to :r = 0. To find g'(0), we need to determine the slope of the line tangent to the graph of g at x = 0. Because this tangent line is horizontal, we conclude that g'(0) = 0. Now, notice the graph off  is linear near x = 0. So f'(0) is just the slope of the line.

Thus, f’ (0) = 2. Substituting this information into h’ (0) gives

h'(0) = .f(0) • g'(0) + g(0) • .f'(0)

=5·0+1-2

=2

(b) To find the derivative of j(x) = g1<-–:cr,we use the Quotient Rule:

.,          (6f(x)) (d (g(:1:) – 7x)) – (g(:1:) – 7x) (d (6f(:1:)))

.7 (:r)=                                  (6.f(:1:))2

(6f(x)) (g'(:1:) – 7x ln(7)) – (g(x) – 7x) (6f'(x)) (6f(x))2

Substituting 3 for x gives

.,( )  _   (6f(3)) (g'(3) – 73 ln(7)) – (g(:3) – 73) (6f'(3))

J   3-                                 (6f(3))2

 

Now, we must find f(3),f'(3),g(3), and g'(3). Looking at Figure 2.4.2, we see f(3) = 3 and g(3) = 2. We also note that g'(3) = 0 because the graph of g is a line with a slope of 0 near :1: = 3. Lastly, .f’ (3) = 0 because the graph of f has a horizontal tangent line at :i: = 3. Substituting these values into j’ (3) gives

.,(.) _ (6(3))(0 – 343ln(7)) – (2 – 343)(6(0))

.7 3 –                            (6(3))2

(18)(-343 ln(7)) – 0

324

-6174ln(7)

:324

343 ln(7)

18

Now that we are able to find the derivative of slightly more complicated functions involving products and quotients, let’s revisit some of the marginal analysis applications we discussed in Section 2.3.

Example 8 Better Purchase, a technology store, has a price-demand function given by p(:1:) = 300(0.997)x, where p(:1:) is the price, in dollars, of each mouse when x mice are sold. Find the marginal revenue when 812 mice are sold, and interpret your answer.

 

Solution: Recall that the marginal revenue is the derivative of the revenue function. Thus, we need to find R'(812). However, first we need to find the revenue function, R, because we were only given the price-demand function, p, in the problem. Remember that

R(:r) = x · p(:1:)

= X (300(0.997r)

= 300:r(0.997r

Thus, R(:1:) = 300:r(0.997r gives the revenue, in dollars, when :T mice are sold. Because R(x) is a product of two functions, we use the Product Rule to find R'(x):

R'(:1:) = (300:1:) (d : ((0.997r)) + ((0.997r) (d :(300x))

= 300:r ((0.997r ln(0.997)) + ((0.997r) (300)

Substituting x = 812 gives

R'(812) = 300(812)(0.997)812 ln(0.997) + 300(0.997)812

;::::; -$37.66 per mouse

This means that when Better Purchase sells 812 mice, their revenue is decreasing at a rate of $:37.66 per mouse.

Recall in the previous section we learned R’ (812) can be used to estimate the revenue from selling the 813th mouse. Thus, we could also say that Better Purchase loses approximately $37.66 in revenue when selling the 813th mouse.

 

 

Average and Marginal Average Functions

Now that we have more tools in our toolbox to help us find derivatives, we can expand upon the ideas we discussed in Section 2.3.

Recall that the average of a group of numbers can be found by adding all the numbers and dividing by how many numbers there are. We can do the same thing to find the average cost of an item: find the total cost of producing all the items and divide by the number of items produced. Likewise, we can define the average cost function to be the cost function, C(:r), divided by the number of items produced, :r. We can define the average revenue function and average profit function similarly:

 

In the previous section, we found the marginal cost, marginal revenue, and marginal profit functions by finding the derivative of the cost, revenue, and profit functions, respectively. We can define similar functions by taking the derivative of the average cost, average revenue, and average profit functions.

 

Remember that a derivative is a rate of change, so the marginal average cost function, for instance, measures the rate of change of the average cost when x items are produced.

Example 9 A company that makes Barbara Dolls has a weekly cost function, in dollars, given by C(:r) = 6000 +

13:r + 0.02×2, where :r is the number of dolls produced.

(a)   Find C (1024) and interpret your answer.

(b)   Find the marginal average cost at a production level of 1024 dolls, and interpret your answer.

Solution:  (a) The average cost function is given by

C(x)=   C(x)=    6000 + l3x + 0.02:r2

:r                      :r

Substituting x = 1024 into C(:r) = 5ooo+i3x+o.o2x2 gives

X

 

2

C:(1024)= 6000 + 13(10  ;:0.02(1024)

40,283.52

1024

:::, $39.34 per doll

 

Thus, when 1024 Barbara dolls are produced, the average cost per doll is $39.34.

(b) Here, we must find the marginal average cost at a production level of 1024 dolls. The marginal average cost is the derivative of the average cost function, so first we need to find C’ (x).

There are two ways we can do this. We can use the Quotient Rule to find the derivative, or we can manipulate the function algebraically and then use the Introductory Derivative Rules. For this example, we will algebraically manipulate the function to find the derivative:

-+-+–

–          6000     13:r     0.02×2

C(x)=

;r           J.:               J.:

f-1, + —

= 6000-1 + 13  ,i.                 0.02×2

l            l

= 6000x-1 +13 +0.02:r

 

Next, we take the derivative:

 

 

 

Substituting J.: = 1024 gives

 

C'(:1:) = -6ooox-2 +o + 0.02

= -6000x-2 + 0.02

 

 

C:'(1024) = -6000(1024)-2 +0.02

:::, -$0.01 per doll per doll

 

When 1024 Barbara Dolls are produced, the average cost per doll is decreasing at a rate of $0.01 per doll.

Example 10 The company in the previous example makes an accessory box for their Barbara Dolls. Their weekly revenue function, in dollars, is given by R(:r) = 500 + 610 ln(:r:), where :r is the number of boxes sold.

(a)   Find R’ (300) and interpret your answer.

(b)   Find .R(300) and interpret your answer.

(c) Find .R’ (300) and interpret your answer.

Solution: (a) Notice that we are asked to find the marginal revenue, not the marginal average revenue. Hence, we take the derivative of the revenue function:

 

R'(:r)=0+610(

-1)

=-610

x         X

 

Substituting :r = 300 into R’ (:r) = 6;0 gives

R'(300) = 610

300

:::::: $2.03 per box

 

Thus, when 300 accessory boxes are sold, revenue is increasing at a rate of $2.03 per box.

(b)   We are asked to find .R(300), which is the average revenue when :r = 300. To find the average revenue function, we divide the revenue function by x:

 

 

 

 

Substituting x = 300 gives

.R(:r)=   .500 + 610ln(x)

:r

 

 

.R(300)=   .500 + 610 ln(300)

300

:::::: $13.26 per box

When 300 boxes are sold, the average revenue per box is $13.26.

(c)   We are asked to find .R'(300), so we need to find the derivative of the average revenue function (i.e., we need to find the marginal average revenue function). Recall that we found the average revenue function in part b:

.R(:r)=   .500 + 610ln(:r)

:r

Unlike the previous example, we will have to use the Quotient Rule to find .R’ (x). Even though there is only one term in the denominator, x, we cannot algebraically manipulate the function and use the Introductory Derivative Rules because of the ln(:r) term in the numerator. Using the Quotient Rule gives

-,          (:r) ( dd (500 + 610 ln(:r))) – (500 + 610 ln(:r)) ( dd· (:r))

R (J:) =         x                                           :r2                                          x

(x) (6;0) – (500 + 610ln(x))(l)

x2

(J:)6}0 -(500+610ln(:r))

x2

610-500- 610ln(x)

x2

110-610ln(:r:)

:r2

Subst1. tutm•

g x = 300 .mto R-‘( x)

 

X 2

= 110-610 ln(x) gi•ves

 

R'(300=)

110 – 610 ln(:300)

(300)2

:::::, -$0.04 per box per box

 

When 300 boxes are sold, the average revenue per box is decreasing at a rate of $0.04 per box.

Try It Answers

1.

8

(a) J'(:r) = x (-1-_) + log (x)

x ln(8)

(b)       R'(t) = t3 ( c1/2 + i) + ( Vf + et) (3t2)

(c)

–t

v'(t) = 7 314

4

– t2

 

(5l ln(t)) – 5t(2t)

2.   (-oo, 0) U (0, oo)

0

,                               (2:r3 + 4:r – 12) (2:r – 3) – (:r2

 

– 3:r + 1) (6:r2

 

+ 4)

‘-‘·  (a) f   (x=)

I ,  _         -14    (¾)

(b) p    (x) –  (ln(:1:))2

(2:r3+ 4x–   12)2

(x3 + 3x) (35×6) –  (5×7) (3×2 + 3x ln(3))

(c)    C(1 :r)=                                    _

(:r3 + 3x)2

4

()  F'(•’) _ ((x-3)(ln(x)))(l2x3-4x+2xln(2))-(3×4-2×2+2x-e)((x-3)(½)+ln(x))

•             a        :I –                                     ((x-3)(ln(x)))2

(b)   s’ (t) = (6t3 + 4t) c5o-ts)( “;:;f•J?     s ; Iog4(t))(-st7)) + (10–;; ;/tl) (18t2 + 4)

 

(c)    f'(t) = (2t8 + 13t2) ( (log3(t)) (14t ln(14)) + 14t (th;(3))) + ((log3(t)) 14t) (16t7 + 26t)

25″‘

•5• Y– _ 6 ,,+.

275

6.     (a) 244 + 64 ln(2)

(b)   -12

7.     (a) -2/49

(b) 7

8.     (a) $26.00 per box; When 250 accessory boxes are produced, cost is increasing at a rate of $26.00 per box.

(b)   $0.04 per box per box; When 250 accessory boxes are produced, the average cost per box is increasing at a rate of $0.04 per box.

Exercises

 

Basic Skills Practice

 

For Exercises 1 – 3, use the Product Rule to find f'(x).

2. J(:r) = (5 + :r) ln(x)

 

For Exercises 4 – 6, use the Quotient Rule to find f'(x).

 

4. f(:r) = x7!:2

 

• 5•

f(n•) _  12x

v.. –             3x+5

 

For Exercises 7 – 10, find the slope of the line tangent to the graph of the function at the given x-value.

 

7.   f(:r) = 8x(:r – 2) at :r = 2

8.   g(:r) =        at :r = -3

9.   h(:r) = Tx””‘t–14 at :r = 4

10. .f(:r) = (ln(:r))(4:r – 18) at :r = 1

 

For Exercises 11 – 13, find the :r-value(s) where the graph off  has a horizontal tangent line.

11. J(x) = 4!t8                                              12. J(x) = ex (x2 – 2x – 7)             13. f(x) = 4×2 ln(x)

 

For Exercises 14-16, x is the number of items made and sold by a company, and the output of the given function is in dollars.

14.   If the cost function is given by C(:r) = 4000 + 32:r + 0.5:r:2, find

(a)   the average cost function, C(:r).

(b)   the marginal average cost function, C’ (;1_:).

15.   If the revenue function is given by R(:r) = 300:1:2(0.997r, find

(a)   the average revenue function, R(:i:).

(b)   the marginal average revenue function, R’ (x).

16.   If the profit function is given by P(x) = 100ft – 600, find

(a)   the average profit function, F(;1_:).

(b)   the marginal average profit function, P’ (:1:).

 

 

 

Intermediate Skills Practice

 

For Exercises 17 – 19, find f'(x).

17.   J(:1:) = 5:1: • 8x + 2×2

18.    f(:1:) = (14J.J + 90:r) (15;1_:3 – log4(;1_:))

For Exercises 20 – 22, find *.

21.  Y = xex

x-7

 

22.       y = ( x :1) (:r50 + 6:r70)

 

For Exercises 23 – 25, find the equation of the line tangent to the graph of the function at the given ;1:-value.

23.     f(:r) = 3t;28 at x = 0               24. g(x) = 15×3 ln(x) at x = l          25. h(x) = x; ;6 at :r = -2

26.   Given f(:r) = 1      l, find the equation of the line tangent to the graph off at :r = 9. Use technology to graph f and the tangent line on the same axes.

 

For Exercises 27 and 28, find the x-value(s) where the graph off  has a horizontal tangent line.

 

27.   J(:r) = ex (:r2 –  2x – 2)

28.   f (X ) = x2-3

 

x2_g

For Exercises 29 and 30, find the :i:-value(s) where f has the given instantaneous rate of change.

29.   f(:r) = x ln(:r); instantaneous rate of change is 1

30.   f(:r) = ;_;g: instantaneous rate of change is 5

31.   Jenn and Barry Ice Cream started an advertising campaign. The company’s sales t months after starting the advertising campaign can be modeled by

–

S(t)=   68t2     952t – 142

t2 –  l4t – 15

thousand dollars. Find a function representing the rate of change of sales t months after the campaign started.

32.   The value of a yacht t years after it is bought is given by

192

V(t) = 4.4 + 3t2+ 50

 

ten thousand dollars. Find V’ (12) and interpret your answer. Round to six decimal places, if necessary.

33.   Sara sells seashells by the seashore. When she sells c shells, her profit is given by

P(c) = vc·(ln(c) – 2) – 10

hundred dollars.  Find the rate of change of profit when Sara sells 40 seashells, and interpret your answer. Round to four decimal places, if necessary.

34.   The company Incensitive makes prank bad smelling incense sticks. The daily revenue function of the company is given by

R(x) = 10×2 • (0.878f

dollars when :r packages of incense sticks are sold.

(a)   Find the company’s marginal revenue function.

(b)   Find R’ (25) and interpret your answer.

(c)   Find the marginal revenue when 18 packages of incense sticks are sold, and interpret your answer.

(d)   Approximate the revenue from selling the 15th package of incense sticks.

(e)   Find the exact revenue from selling the 12th package of incense sticks.

(f)   Estimate the revenue if 30 packages of incense sticks are sold.

(g)   Find the exact revenue if 18 packages of incense sticks are sold.

 

35.   Forry is a videogame company that makes the Gamestation console. The company’s weekly cost function for making :r thousand consoles is given by

 

C(:r) –

2300:1:3 + 90:r2 + 21250

3

•      –                :r  + 125

thousand dollars. Find the marginal cost function for Fony’s consoles.

36.   An adult blue dragon is observed flying over a city. Its position, in feet, is given by

s(t=)  40t2 + 40t + 67

t2 + 1

where t is the number of seconds after it passes overhead. Find a function v representing the velocity of the dragon t seconds after it passes overhead.

37.   The table below shows values of f(:r), g(:r), f'(:r), and g'(:r) for certain values of :r. Use the information in the table to find each of the following.

 

X	f(x)	g(x)	f'(x)	g'(x)
13	16	15	13	13
14	15	15	14	15
15	17	17	14	14
16	16	13	15	13
17	15	13	14	14

(a)    h'(15), where h(x) = f(x) • g(x)

(b)    j'(l3), where j(x) = f(;t;t

(c)   k'(16), where k(x) =  g(;l·! x)

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

(a)    h'(-5), where h(:r) = g(:r)f(x)

(b)    j'(8), where j(:r) = !J;l – g&)

 

39.   A company that makes clothes dryers has a weekly cost function, in dollars, given by

C(:r) = 3000 + 12:i: + 0.2J:2

where x is the number of dryers produced.

 

(a)   Find C(323) and interpret your answer.

(b)   Find the marginal average cost at a production level of 227 dryers, and interpret your answer.

(c)   Find the exact cost of producing 933 dryers.

40.   Isaac Ay makes build-it-yourself desks. He has a revenue function, in dollars, given by

where x is the number of desks sold.

(a)   Find .R(60) and interpret your answer.

(b)   Find the marginal average revenue when 237 desks are sold, and interpret your answer.

(c) Find the exact revenue from selling 97 desks.

41.   A company that makes designer hairbrushes has a weekly profit function, in dollars, of

P(:1:) = 410 – 11:I: – 0.4:1:2 + 610 ln(:1:)

where xis the number of hairbrushes made and sold.

(a)   Find P(20) and interpret your answer.

(b)   Find the marginal average profit when 16 hairbrushes are made and sold, and interpret your answer.

(c)   Find the exact profit from making and selling 30 hairbrushes.

 

 

 

Mastery Practice

 

For Exercises 42 – 46, find f 1(x).

42.    f (:1:) = ( fa – 18:1:) · 2(10?

 

 

45.    f(x) = (4×3 –  8:1:)-1 (18 (0.5x) + 37)

43•

j‘(:,I•,)- – ( ln(!xJ)x+31+91xG ) ( eX – 7)

 

 

46.   j(x) =( a,-,,

 

 

 

4x)lf4+7r3

vx2+8ex (2log(x)-9.7)

 

47.  The table below shows values of f(x), g(x), f'(x), and g'(x) for certain values of x. Use the information in the table to find each of the following.

 

;r	f(x)	g(:1:)	f'(x)	g'(x)
-2	1	0	-2	-2
-1	0	0	-1	0
0	2	2	-1	-1
1	1	-2	0	-2
2	0	-2	-1	1

a)

(

h'(l) where h(:1:) =  f(x)-g(x)

‘                                  J(x)+g(x)

(b)   j'(-2), where j(x) = f(:1:) • (8:r – g(:r))

(c)   k'(l), where k(:1:) = 3g(x) • 4x

For Exercises 48 and 49, find the equation of the line tangent to the graph of the function at the given x-value. Use technology to graph the function and the tangent line on the same axes.

49. h(:i:) =  x2+13x 8 at J: = 4

x+5vx

 

For Exercises 50 and 51, find the :r-value(s) where the graph of the function has a horizontal tangent line.

.SO. f(:r) = 8!;;:8                                                                                       -51. g(x) = ex (:r2 – 3) – 31

 

For Exercises 52 and 53, find the :r-value(s) where the graph of the function has the indicated slope.

52. f(x) = ex (J’.2 – 6x – 15) – 8x; slope is -8

.53. g(:r•) — x3––4x’• slope i’s – l9

54. A snowball is thrown with a velocity given by

1.87t2 + 4t + 4.83

Vt( ) =          t2 + 1

feet per second t seconds after being thrown. At what rate is the snowball decelerating 2 seconds after it is thrown? Round your answer to three decimal places, if necessary.

5-5. Mr. Wutz owns a hot dog business named Wutz UpDogs. The daily profit function when x hot dogs are sold is given by

dollars. Calculate P'(20) and interpret your answer. 56. Mr. Wutz, the owner of Wutz UpDogs, started an advertising campaign. He found that his sales, in dollars, t weeks after the campaign started is given by

S(t) =

4t + 7)

10( (1.2)t

Find the rate of change of sales two weeks after the campaign started.

56.   A company has determined their yearly cost function is given by

C(x) = 500 + 610ln(x)

dollars when x pounds of product are made. Find the marginal average cost when 300 pounds of product are made, and interpret your answer.

57.   The general store Nookington’s has a department selling shovels. Tom Nook, the proprietor, determines the department’s monthly revenue function is

R(J.:) = 24x – 5J-.: log(x)

dollars, where :r is the number of shovels sold.

(a)   Find the marginal revenue when 50 shovels are sold, and interpret your answer.

(b)   Approximate the revenue from selling the 50th shovel.

(c) Find the exact revenue from selling the 50th shovel.

(d) Estimate the revenue if ,50 shovels are sold.

(e) Find the exact revenue if 50 shovels are sold.

58.   A company that makes and sells dishwashers has a monthly profit function, in dollars, given by

where :r is the number of dishwashers made and sold.

(a)   Find P(234) and interpret your answer.

 

(b)   Find the marginal average profit when 209 dishwashers are made and sold, and interpret your answer.

(c)   Find the exact profit from making and selling 157 dishwashers.

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

(a) h‘(– 1), where, h (x) — f(x))+–gg((x)

(b) k'(5), where k(:r) = x2 f(x) + J.:3g(x)

 

 

 

Communication Practice

 

61. Does the derivative of a product of functions equal the product of their derivatives? Explain.

v           oes dx       5x )

“2. D ,    ..!i.. ( 3x4                         —11i:_?  E    1 •

—  5x ln(6) •       xp am.

63. Kathryn was helping her classmate, Vanessa, with one of her calculus homework problems. The problem said:

3

–

“F’ d f'(· ·) ·t· f( ·=)  ln(J.:-) nJ.:      4 ”

Ill            J.  1        J.

6ex – :r 2         .

Kathryn told Vanessa all she has to do is use the Quotient Rule. She wrote the following on Vanessa’s paper:

TB’-BT’

B2

 

If Vanessa uses the rule Kathryn wrote on her paper, will she get the correct answer? Explain.

2.5        TheChainRuleWe have seen a wide variety of techniques for calculating the derivative of a function, but there are still many functions for which our rules will not apply. For some functions, we can algebraically manipulate the function first and then apply our previous rules, but the amount of algebraic manipulation involved becomes unrealistic.

For instance, the derivative of y = (3×2 + 1)99 can be found by first multiplying (3×2 + 1) by itself ninety­ nine times, but this is not reasonable as this algebra would take hours, if not days, to perform by hand. Even though it is possible to find the derivative that way, it is not in any way reasonable.

 

But for some functions, we currently have no method, reasonable or unreasonable, which will allow us to find the derivative. For instance, we do not currently have a technique that would allow us to find the derivative of y = J3:r2 + l. Notice, though, we can write this function as a composite function of the form

y = f(g(:r)), where f(x) = .Ji and g(x) = 3:r2 + 1, and we can use the Introductory Derivative Rules to

find the derivatives of both f and g!

 

Our focus in this section is to find derivatives of functions that are compositions of functions. We will learn a new rule, called the Chain Rule, which will allow us to do just that!

 

Learning Objectives:

In this section, you will learn how to use the Chain Rule to calculate derivatives of a composite functions, as well as functions that have not yet been composed, and solve problems involving real-world applications. Upon completion you will be able to:

•   Calculate the derivative of a composite function using the Chain Rule.

•   Calculate the derivative of composite function before it has been composed using the Alternate Form of the Chain Rule.

•   Calculate the derivative of logarithmic functions by first expanding them using the Properties of Log- arithms.

•   Calculate the slope of a tangent line using the Chain Rule.

•   Find the equation of a tangent line using the Chain Rule.

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

•   Determine the x-value(s) where the graph of a function has a horizontal tangent line using the Chain Rule.

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

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

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

•   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 Chain Rule.

•   Interpret the meaning of marginal cost, revenue, and profit.

•   Estimate the cost, revenue, or profit of an item using marginal analysis and the Chain Rule.

•   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 Chain Rule.

•   Calculate marginal average cost, revenue, and profit using the Chain Rule.

•   Interpret the meaning of marginal average cost, revenue, and profit.

To help us develop the Chain Rule, let’s look at the derivative of two basic, yet similar, functions. First, consider the functionF1(J.:) = (:fr2 + 1)2. We can use the Introductory Derivative Rules to find the derivative of Fi by first algebraically manipulating the function:

F1(:1:) = (3:1:2 + 1)2

= (3×2 + 1) (3:1:2 + 1)

= 9×4 + 6×2 + 1

Because this function is now written as a sum of basic functions, we can easily find the derivative using the Introductory Derivative Rules:

F{(x) = 36×3 + 12x

We will rewrite this derivative slightly to help with the development of our new rule:

Fi'(x) = 36×3 + 12x

= 12J.: (3J.:2 + 1)

= 2 (3:1:2 + 1) (6x)

At this point, you may be able to see how to find the derivative ofF1 without first expanding the function, but before we discuss it, let’s look at a second, similar function: F2(x) = (3:1:2 + 1)3. Notice the only difference betweenF2 and Fi is now the function (3:1:2 + 1) is being raised to the third power instead of the second power. Again, we first algebraically manipulate the function:

F2(:1:) = (3:1:2 + 1)3

= (3J.:2 + 1) (3:r2 + 1) (3×2 + 1)

= (3×2 + 1) (9×4 + 6×2 + 1)

= 27J.:6 + 18J.:4 + 3×2 + 9×4 + 6J.:2 + 1

= 27:1:6 + 27:i:4 + 9:1:2 + 1

Now, we can easily find the derivativeF2′(J.:). Again, we will algebraically manipulate the result so it is in our desired form:

F2′(x) = 162×5 + 108:1:3 + l8x

= 18:r (9J.:4 + 6J.:2 + 1)

= l8x (3×2 + 1) (3×2 + 1)

= 18:1: (3×2 + 1)2

= 3 (3:r:2 + 1) 2 ( 6J:)

Our results are summarized below:

Fi(x) = (3×2 + 1)2 andF{(:1:) = 2 (3:1:2 + 1) (6x)

F2(x) = (3×2 + 1)3 andF2′(x) = 3 (3:1:2 + 1)2 (6x)

Notice that for both of these functions, to find the derivative without algebraically manipulating the original function first, all we have to do is bring the power down to the front, leave the function on the inside, 3:1:2 +1, alone, subtract one from the original power, and then multiply by the derivative of the “inside” function, 3×2 + 1. This is a special case of the Chain Rule that we will discuss in more detail in this section.

 

For now, notice both Fi and F2 are composite functions of the form f(g(:r)), where g(:r) = 3×2 + 1 and f (x) = :r2 for F1 (:r) and f (;r) = :r3 for F2(:r). Thus, to find the derivative of F1 and F2, we found the derivative of the “outside” function, f, evaluated at the “inside” function, g, and then we multiplied by the

derivative of the “inside” function, g. This is exactly what the Chain Rule tells us to do, and it is summarized below.

 

 

Using the Chain Rule

In the following example, we will use the Chain Rule to find the derivative of each of the functions discussed in the introduction.

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

(a)   Fi(:1:) = (3:1:2 + 1)2

(b)  F2(:1:) = (3:1:2 + 1)3

(c) F3(:r) = (3:r2 + 1)99

(d)   F4(:1:) = J3x2 + 1

Solution:  (a) Because F1 can be written as the composite function F1(:1:) = f(g(x)), where f(x) = x2 and

g(x) = 3×2 + 1, we can use the Chain Rule to find Fi'(x):

Fi'(x) = J'(g(x)) · g'(x)

Using the Introductory Derivative Rules, we know f'(:1:) = 2:1:, Thus,

J'(g(x)) = J’ (3×2 + 1)

= 2 (3:1:2 + 1)

Again, using the Introductory Derivative Rules, we know g'(:r) = 6x. Substituting this into the Chain Rule gives

Fi'(x) = J'(g(x)) · g'(x)

= 2 (:3:1:2 +1) (6x)

Notice this is the same answer we reached previously!

(b) F2(:r) = (3:r2 + 1)3 can be written as the composite functionF2(:r) = f(g(:r)), where f(:r) = :r3

and g(x) = 3×2 + 1. Thus, the Chain Rule can be used to findF2′(x):

Again, the Introductory Derivative Rules give us f'(:1:) = 3:1:2 and g'(:1:) = 6:1:, Thus,

J'(g(:r)) = J’ (3:r2 +1)

= 3 (:3:1:2 +1)2

Substituting this into the Chain Rule gives

F2′(:1:) = J'(g(:r)) • g'(x)

= 3 (3:r2 + 1) 2 ( 6:r)

 

Again, notice this is the same answer we reached previously!

(c)   F3(x) = (3×2 + 1)99 can be written as the composite function F3(x) = J(g(x)), where J(x) = x99

and g(x) = 3×2 + l. Thus, the Chain Rule can be used to find F3′(x):

The Introductory Derivative Rules give us f'(x) = 99:r98 and g'(:r) = 6:r. Thus,

J'(g(x)) = J’ (3×2 + 1)

= 99 (3×2 + 1)98

Substituting this into the Chain Rule gives

F31(x) = J'(g(:r)) • g'(:r)

= 99 (3×2 + 1)98 (6x)

Using the Chain Rule is much easier than expanding the original function like we discussed at the beginning of this section!

(d)   Although F4(x) =3:1:2 + 1 might appear to be very different than the other functions, if we rewrite the radical as a power function, we see that the same ideas hold:

F4(x) = J3x2 + 1 = (3:r2 + 1)½

Thus,F4(:r) can be written as the composite function F4(x) = f(g(x)), where f(:r) = x112 and

g(x) = 3×2 + l. Thus, the Chain Rule can be used to find F4′(x):

F41(:r:) = J'(g(:r:)) • g'(:r:)

Again the Introductory Derivative Rules give us f'(x) = ½x-1/2 and g'(x) = 6x. Thus,

J'(g(:r)) = J’ (3:r2 + 1)

= i (3:r2 + 1– ) ½

Substituting this into the Chain Rule gives

F41(x) = J'(g(x)) • g'(x)

=

2

1 ( 3x2 + 1)_.2! ( 6x)

 

 

 

Generalized Power Rule

The functions in the previous example all represent a specific case of the Chain Rule where the “outside” function is a power function. This specific case is summarized below.

 

Note: In the above theorem, if g(x) = :r, then we would have d  (:rn) = n:rn- l ( ix  (:r)) = nxn-l (1) = n:1:n-l.

Notice this is the Power Rule we know and love from Section 2.3! Let’s get more practice using the Chain Rule!

 

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

(a)    y =l-:r

b)

(    Y  –

1  –     (2xs+15)8

ln(x)

(c) y=  e8-4xa

Solution:  (a) We begin by rewriting the function as y = (1 – :i:)112. The Chain Rule, or more specifically, the Generalized Power Rule, gives

 

y’ = !(1 – a:)½-1 ( (1- :r))

2                              da:

= 1 (1 – x)-2l (-1)

2= – 1(1 – x)-21

2

(b)

1

The most “outside” part of y = ( 2fi5c!i5 )8

 

is a power, so we use the Generalized Power Rule:

 

y=‘

 

8 1

5

8 (2x +15) –

 

(          (2a: + 15))

5

ln(x)               da:      ln(x)

To find the derivative of the “inside” function, we need to use the Quotient Rule:

!!:._ (2×5 + 15)=     (ln(:r)) (lx (2:r5 + 15)) – (2:r5 + 15) (d (ln(:r)))

dx       ln(a:)                                        (ln(a:))2

(ln(:r)) (10:r4) –  (2a:5 + 15) (¾)

(ln(a:))2

Now, we substitute this result into the derivative to get the final answer:

5

8 1

,        8 (2x +15) –

 

( d    (2x + 15))

5

y =           ln(x)               da:      ln(x)

(2×5 +

7

=                  15) ((ln(x)) (10a:4-)     (2a:5 + 15) (¾))

8      ln(x)                            (ln(x))2

(c) Notice y = e8-4x3 has an x in the exponent. Thus, we are no longer dealing with a power function, but an exponential function instead. Therefore, we return to the general form of the Chain Rule to find the derivative:

!!:._ (f (g(x))) = J’ (g(x)) · g’ (x)

dx

Because the Chain Rule allows us to find the derivative of a composition of two functions, we need to be able to identify the two functions. We have seen ex in previous sections, but now we have a more involved function of a: where we typically see just :i:. Thus, the “outside” function is J(x) = ex and the “inside” function is g(a:) = 8 – 4a:3 as shown below:

f(g(x)) = f (8 – 4a:3)

=   e8-4x3

 

Because f’ (x) = ex, we have

 

 

 

We also need the derivative of g:

 

J'(g(x)) = J’ (8 – 4×3)

= e8-4x3

 

 

g'(:r) = -12×2

 

Substituting all of these pieces into the Chain Rule gives

y’ = .f'(g(:r)) • g1(:r)

= e8–4×3 (-12×2)

=  -12x2e8-4xs

 

 

Try  It 1

 

 

Generalized Exponential (base e) Rule

The solution to part c of the previous example leads us to another specific case of the Chain Rule where the “outside” function is .f (:1:) = ex. We refer to this rule as the Generalized Exponential (base e) Rule!

Theorem 7 The Generalized Exponential (base e) Rule

 

Note: In the above theorem, if g(:r) = :r:, then we would have d  (ex)= ex (lx(:r:)) = ex(l) = ex. Notice this is the Exponential (base e) Rule we know and love from Section 2.3!

Let’s get more practice using the Chain Rule!

 

Example 3 Find the derivative of each of the following functions.

(a)    y = e7+2x4–fi

(b)   .f(x) = (e3-¾) (:r-4)

(c)   C(:r) = 413×2-12x

 

Solution:  (a) Here, we have a composition of two functions where the “outside” function is ex. Thus, we can use the Generalized Exponential (base e) Rule:

y’ = _!!_ (e7+2×4-fi)

dx

= e7+2x•–fi  ( d! (7 + 2:.r4 – v’x))

 

= e7+2×4-fi ( d! (7 + 2:r:4 – :1J))

 

= e7+2x4-fi ( 8:1:3 – x-½)

(b) Because f (:r) = (e3-¾) (:r – 4) is a product of two functions, we start with the Product Rule:

 

We can use the Introductory Derivative Rules to find the the derivative of J.: – 4 : d (J.: – 4) = 1, but we must use the Generalized Exponential (base e) Rule to find the derivative of e3-¾:

 

–d    ( e3_ 2x ) -e 3_2x ( –d    ( 3– 2 ) )

dx                            dJ.:        J.:

2x- ))

= e3-¾ (d : (3 –       1

=ex 3–‘- (2x-2)

 

Substituting both derivatives into f’ (:r) gives

J'(x) = (e3-¾) (d (x-4)) +(x-4) (d  (e3-¾))

= (e3-¾) (1) + (:.r –  4) ( e3-¾(2J.:-2))

= e3-¾+ 2x-2(x – 4)e3-¾

(c) C(x) =  13    12

2

4  x –  x is not of a form in which we can use the Generalized Exponential (base e) Rule,

so we will have to go back to the general form for the Chain Rule. C(J.:) can be viewed as the composition function, f(g(x)), where f(:.r) = 4x and g(;r) = 13:.r2 – 12x.

Because f'(:r) = 4x ln(4), we have

J'(g(:r)) = J’ (13:i:2 -12:r)

= 413×2-12x(ln(4))

 

Next, we need to find g'(x):

g'(J.:) = 26:1; -12

Substituting these pieces into the general form of the Chain Rule gives

C'(;i:) = J'(g(:r)) · g'(J.:)

–

= 413×2 12x(ln(4))(26x -12)

 

 

Generalized Exponential (base b) Rule

The solution to part c of the previous example leads us to an even more generalized exponential derivative rule that applies to exponential functions of any base: the Generalized Exponential (base b) Rule!

 

Theorem 8 The Generalized Exponential (base b) Rule

 

 

 

 

 

 

 

Note: In the above theorem, if g(x) = x, then we would have lx (bx)= bx(ln(b)) (d  (x)) = bx(ln(b))(l) =

bx ln(b).

Notice this is the other Exponential Rule we know and love from Section 2.3! Notice also that this formula gives the same result as the Generalized Exponential (base e) Rule when b = e because ln(e) = l.

Let’s get more practice using the Chain Rule!

Example 4  Find       for each of the following functions.

(a)    y=  5sx-22×10

(b)    y = 10  x2 + 7

(c)   y = 21n (r7 – :1:t)

Solution:  (a) This function is of a form in which we can use the Generalized Exponential (base b) Rule. Here,

b = 6 and g(x) = gx -22:i::10, so

 

dy = 5sx-22x”‘ (ln(6)) (_!!_ (sx – 22×10))

dx                               dx

= 5sx-22x”‘ (ln(6)) (8x(ln(8)) – 220×9)

(b) The “outside” part of y = 10{1/x2+7 is 10 raised to a power, so we start with the Generalized Exponential Rule with base 10:

dy = l0{1/x2+7(1n(10)) (_!!_ ({/r2 + 1))

d:r                                d:r

= 10{1/x2+7(ln(l0)) ( d  ( (:1:2 + 7)½))

 

Next, we apply the Generalized Power Rule to find lx ( (:1:2 + 7) ½):

Substituting this into the derivative gives

 

dy = 10{1/x2+7(ln(l0)) (_!!_ ((x2 + 7) ½))

dx                                dx

= 10{1/x2+7(ln(10)) (    (x2 +7)-  (2x))

 

(c) For the function y = 2 ln ( x 7 – :d), the previous case of the Chain Rule does not apply. Once again, we must go back and use the general form of the Chain Rule. Perhaps you see the pattern by now and notice that we will get a rule for natural logarithms next!

Using the parlance of the general Chain Rule, we let f(x) = ln(:1:) and g(x) = x7 – xi. You may be curious about the 2 in the original function. Remember that constants come along for the ride, so we will find the derivative of h(:r) = f (g(:r)) = ln (r7 – ;rt), and afterwards, we will multiply by 2 to get the final answer.

 

Because f’ (x) = ¾, we have

 

f’ (g(:r)) = J’ (:r:7 – :r:i)

1

 

 

 

Next, we need to find g'(:i:):

 

 

Therefore,

 

g‘( x)

 

= 7×6 – 1-:r-“6

7

h'(:r) = J'(g(:r)) • g'(:r)

=      1      (7×6 –  !x-¥)

:1.:1-:rt               1

7:1.:6 –       x-¥

x7 -xi

Multiplying by the constant 2 gives the final answer:

 

 

 

Generalized Natural Logarithm Rule

The solution to part c of the previous example leads us to another specific case of the Chain Rule in which the “outside” function is f(x) = ln(x). We refer to this rule as the Generalized Natural Logarithm Rule!

 

0

Note: In the above theorem, if g(x) = :1.:, then we would have d (ln(x)) = ¾ U, (:1.:)) = ¾(1) = ¾- Notice

this is the Natural Logarithm Rule we know and love from Section 2.3!! Let’s get more practice using the Chain Rule!

Example 5 Differentiate each of the following functions.

(a)    y = ln (x2 +:r – x-1 – :r-2)

(b)    f(t) = ln Ct2 1)5)

 

(c)   y=log((4:r2-12:r)9)

Solution: (a) If we let g(:r) = x2 + :r – x-1 – x-2, then we can use the Generalized Natural Logarithm Rule.

Because g'(x) = 2x + 1 + x-2 + 2x-3, we have

 

t

)” ) , there are two ways we can find f’ (

2  1

 

(b)   For f (t)=ln((

t). One option is to use the Generalized

Natural Logarithm Rule, then the Quotient Rule to find the derivative of the “inside” function, and then the Generalized Power Rule to find the derivative of the numerator of the “inside” function.

However, we have another option: algebraically manipulate the function first using the Properties of Logarithms to make it easier to find its derivative. We will demonstrate the second option because it is much easier. Rewriting the function using the Properties of Logarithms gives

 

f (t=)

ln ( ( t2 ; 1)5 )

 

= ln ( (t2 + 1) 5) –  ln (i)

= 5 ln (t2 + 1) – t

 

Even before we start taking the derivative, we can see how this technique is easier. We still need to use the Generalized Natural Logarithm Rule, but we eliminated the need for the Quotient Rule and Generalized Power Rule:

J’ (t) = .5 ( :t  (ln (t2 + 1))) – (! (t))

–  .5(  ft ( t2 + 1))    -1

t2 + 1

 

=5 (

t2+2t  )     1

1  –

lOt

=2—1

t + 1

Note: As shown in this example, using the Properties of Logarithms to algebraically manipulate the function first often reduces the need for more complicated derivative rules. You do have the tools to attempt to find the derivative directly, and doing so should make you appreciate the approach we took here.

(c)   Recall there are only two instances where we do not have to write the base of a logarithm. The natural logarithm, ln(:r), implies a base of e, and the common logarithm, log(:r), implies a base

of 10. Here, for the function y = log ( (4×2 – 12x) 9), we have the second case. Thus, we cannot

use the Generalized Natural Logarithm Rule. As you probably expect by now, this solution will lead us to the Generalized Logarithm (base b) Rule!

Before we find the derivative, let’s first apply the Properties of Logarithms to rewrite the function:

 

y = log ( (4:r2 –  12:.r/)

= 9 log (4:r2 –  12:r)

 

Notice we are dealing with a composition of functions, and thus, we need to use the general form of the Chain Rule to find the derivative. The “outside” function is f (:r) = 9 log(a;), and the “inside” function is g(x) = 4×2 – 12x.

Because f’ (x) = 9 • ½ · ln( O), we have

J'(g(x)) = J’ (4×2 -12x)

= 9.

1             1

4×2 – 12:r•  ln(lO)

9

 

 

(4J:2 – 12:1:) (ln(lO))

Next, we need to find g'(x). Recalling g(x) = 4×2 – 12x gives

g’ (:r) = 8:r – 12

Therefore, recalling f'(g(:r)) =(4x2_12;)(ln(l0)), we have

y’ = J'(g(x)) · g'(:r)

= (  (4:r2

9                )

–  12:r) (ln(lO))

(8•:r – 12)

9(8:r – 12) (4×2 -12x) (ln(lO))

72:r – 108

(4:r2 -12:r) (ln(lO))

 

 

Generalized Logarithm (base b) Rule

The general form of the rule from part c of the previous example is

 

Note: In the above theorem, if g(:r) = :r, then we would have ix (logb(:r)) = ½ ln(b) (ix  (x)) = ½ ln(b) (1) =

xl (b). Notice this is the Logarithm (base b) Rule we know and love from Section 2.3!! Let’s get more practicing using the Chain Rule!

Example 6 Find the derivative of each of the following functions.

(a)   H(:r) = log18 (:rJ – 22:r5 + 55:1:½+ 52:r2)

(b)    y = (log (x3 – 187)) (log3 (4:r5 – 1))

(c)   F(x) = log5 (c5x”-; :(x+l))

(d)   B(t) = log ( •( c J ;J; )8)

Solution:  (a) The “inside” function here is g(:r) = x    – 22×5 + 55x½ + 52×2, so

I                5 23                               4         5.5    1

g (:r) = ir      – llOx +2x-2 +104x

Using the Generalized Logarithm Rule with base 18, we have

l:!2x -110×4 + 525x-½+ 104:r

H'(:r) =

 

 

(:1J – 22×5 + 55x½ + 52×2) (ln(18))

(b)   As we have seen in previous examples, we often have to apply more than one derivative rule to a function in order to find its derivative. For instance, the most “outside” part of this function,

y = (log (x3 – 187)) (log3 (4×5 – 1)), is multiplication, so we need to start with the Product Rule. Then, to find the derivative of each factor, we will use the Generalized Logarithm Rule (first with base 10 and then with base 3):

y’ = (log (:1:3 -187)) (d  (log3 (4×5 -1))) + (log3 (4:1:5 –1)) (d : (log (x3 -187)))

_4_ (4:1:5 -1) )                   .           (       _4_ (:1:3 -187) )

= (log(:r:-3187)) ( (4: -1) (ln(3))    + (log3 (4×5-   1))    (x3d 187) (ln(lO))

3 87

4

(         20:r           )                4

2

5          (

3:1:            )

=(log(:1:-l

)) (4×5-l)(ln(3)) +(log3(

x  -l))  (:r3-187)(ln(10))

20×4 log (x3 – 187)      3×2 log3 (4x5 – 1)

=    (4×5 -1) (ln(3)) + (x3 – 187) (ln(lO))

(c)   F(x) = log5 (c5x3_; :(x+l)) is a function that we could immediately start differentiating using the derivative rules, but we will save time and effort by using the Properties of Logarithms to rewrite the function first:

F(x)=  log5(  (5×3-    :; (:r+ 1))

3        2

2

= log5 ( 25x ) – log5 ( (5:1: – 10) (x + 1))

5                                        5

= log5 ( 25×2) –             (log (5:i:3 – 10) 2 + log (x + 1))

= log5(

25x~”)

– log5 (5×3 – 10)2

– log5(:r + 1)

= :r-2 log5(25) – 2log5 (5:r3 – 10) – log5(:1: + 1)

= 2:r:2 –  2log5 (5×3 -10) – log5(x + 1)

Taking the derivative gives

F’( X)=    4x–    2 (  lx ( 5:1:-3    10) ) –         ix (:1:+1)

3

<                                  (5x – 10) (ln(5))         (:1: + l)(ln(5))

– 4x–  2 (         15:1:2        ) –             1

3

–                <                                  (-5x  -10)(ln(5))         (x+l)(ln(-5))

30×2                                     1

=4x- (5J:3 -10) (ln(5))     (:1: + l)(ln(5))

(d)   To find the derivative of B(t) = log ( •(c J ;;1)8), we will use the Properties of Logarithms to rewrite the function first:

4       3            8

B(t) = log

4t 1-Q(3t +l2) )

((

 

= 8 log(

5t8_27

4t4 – 3t3 + 2)

1Q(5ts-21)

t –

= 8 (log (4t4 – 3t3 + 2) – log ( 10(5 3 27)))

10               t –

= 8 (log (4t4 – 3t3 + 2) – log    ( 10(5 8 27)))

= 8 (log (4t4 – 3t3 + 2) – (5t8 –  27))

= Slog (4t4 – :3t3 + 2) – 8 (5t8 –  27)

= Slog (4t4 – 3t3 + 2) – 40t8 + 216

Now, to take the derivative, we will use the Generalized Logarithm Rule with base 10:

–

I         _  (          -it (4t4       3t3 + 2)    )       ,    _T

B  (t) =8     (4t4 – 3t3+ 2) (ln(lO)) –     320t +0

–

8             16t3

9t2         )

32  7

=     (  (4t4 – 3t3 + 2) (ln(lO)) –        0t

128t3 –  72t2

(4t4 – 3t3 + 2) (ln(lO)) – 320t7

 

All of the specific cases of the Chain Rule we have encountered are summarized below.

Applications

Example 7 Find the equation of the line tangent to the graph of the following function, f, at:r = 1:

j(J:) =

ln(2 – :r)

9×2 – 5×3

Solution: Remember that in order to find the equation of any line, we need its slope and a point on the line. To find the point for this line, we calculate f(l):

.              ln(2 – 1)          ln(l)

j(l)= J9(1)2- 5(1)3=    y’4 = O

 

Thus, the tangent line passes through the point (1, 0).

J;;;=

Next, we need the slope of the tangent line. To find the slope of the tangent line, we calculate f'(l). But first, we must find J'(:i:). Because the outermost operation of f(:r) = 3 is division of two functions, we begin with the Quotient Rule:

J’(a:)= 9×2 – 5×3 Ux (ln(2 – :r))) – (ln(2    a:)) U{ 9( x2 – 5×3))

9( x2 – 5×3)

9×2 – 5×3 (ix(ln(2-:r))) – (ln(2 – x)) (ix ((9×2 –  5×3) ½))

9J:2 – fo:3

To find d  (ln(2 – J:)), we use the Chain Rule where the “inside” function is g(J:) = 2 – J:, making

g'(x) = -1. This yields

rl                     g'(a:)

rlx(ln(2-:r))=  g(x)

-1

2 -a:

To find ix ( (9a:2 –  fo:3)112), the “inside” function is g(x) = 9×2 –  5:r3, making g'(a:) = 18x – lfo:2. This yields

 

Substituting these derivatives into f’ (:r) gives

9×2 – 5×3 (…4..(ln(2 – :r))) – (ln(2 – a:)) (…4.. ((9×2 – 5×3) ½))

f'(a:)=                                dx                                                              dx                                       

•                                                                                                                                                       9:r2 –  5:r3

9×2 – 5×3 (2-:::_ ) – (ln(2 -:r)) (½ (9×2 – 5×3)-½(18a: – lfo:2))

9×2 –  5:r3

Now, we can calculate J'(l) to get the slope:

,     _       J9(1)2 – 5(1)3 ( 2-:::_11) –  (ln(2 – 1)) ( ½ (9(1)2 – 5(1)3)-½(18(1) – 1.5(1)2))

 

f ( 1)-                                                   9(1)2 – 5(1)3

v’4(-1) – ln(l) ( ½ · 4-½· 3) 4

-2-0

4

-2         1

4         2

 

Using the point-slope form of the equation of a line gives

 

 

 

Example 8 Table 2.10 shows values of f(:r),g(:r),J'(:r), and g'(:r) for certain values of :r. Use the information in the table to answer each of the following.

 

:r	.f (:r)	g(:r)	.f'(:r)	g'(:r)
1	2	1	3	4
2	3	4	1	2
3	1	3	3	3
4	3	2	5	2

 

Table 2.10: Various values of f(:r), g(:r), f'(x), and g'(x)

(a)   If w(:r) = g(.f(2:r:)), find w'(l).

(b)   If n(:r) = 3:-r log9(f(:r)), find n'(3).

Solution:  (a) The function w(x) = g(f(2x)) is a composition of three functions, so we will apply the Chain Rule two times.

First, we have

w'(:r) =

d

g'(f(2J.:)) • -(!(2:r))

dx

Applying the Chain Rule again to find the derivative of d  (f(2x)) gives

 

.!},_(f(2J.:)) = J'(2x) (.!},_(2x))

dx                            dx

= J'(2x) • 2

 

Therefore,

w'(J.:) = g'(f(2J.:)) • .!},_(!(2:r))

dx

= g’ (f (2J.:)) · J’ (2J.:) · 2

Lastly, we substitute x = l and use the relevant function values from Table 2.10:

w'(l) = g'(f(2)) · J'(2) · 2

= g'(3) • 1 · 2

= 3 -1 • 2

=6

(b) For n( x) = 3x– log9(f (x)), the outermost operation is multiplication of two functions. So we start with the Product Rule:

To find ix (log9 (.f (x))), we use the Generalized Logarithm Rule with base 9:

d                              f'(:r)

dx (logg(.f(:r))) = (.f(x))(ln(9))

 

Substituting this derivative, we have

 

n'(:r) = 3:r • d  (log9(.f(x))) + (log9(.f(:r))) • 3

.           .f'(:r)                               .,

= 3x • (.f(:r))(ln(g)) +(log9(.f(x))). ‘-‘

 

Substituting :r = 3 and using the relevant function values from the table gives

/  .                     !'(3)

n (3) = 3(3) • (.f(3))(ln(g)) + (log9(.f(3))) • 3

3

= 9 • l(ln(g)) + (log9(1)) • 3

27

= ln(9)+ O

27

ln(9)

 

 

Example 9 The graphs of the functions f and g are shown in Figure 2.5.1. Use the graphs to answer each of the following.

 

Figure 2.5.1: Graphs of the functions  f and g

(a)   If h(:i:) = f(g(J:)), find h'(-3).

(b)   If j(J:) = ef(x)·g(x2), find j'(O).

 

Solution: (a) For the function h(:r) = f(g(x)), the Chain Rule tells us h'(x) = f'(g(x)) • g'(x). Therefore, h'(-3) = f'(g(-3)) • g'(-3). To calculate h'(-3), we will first find g(-3). Looking at the graph directly, we see that g(-3) = 6. This means J'(g(-3)) = f'(6). So now we know

h'(3) = J'(g(-3)) · g'(-3)

= J'(6) · g'(-3)

Next, we must find f'(6) and g'(-3)..f'(6) is the slope of the graph of .f at :r = 6, and g'(-3) is the slope of the graph of g at :r = -3. Because both graphs are linear near these x-values, can calculate their slopes (rise over run). We see f'(6) = 2 and g'(-3) = -1.

Thus,

h'(-3) = J'(g(-3)) · g'(-3)

= J'(6) • g'(-3)

= (2)(-1)

= -2

(b) To find :i'(:r), where j(:r) = ef(x)·g(x2), we start with the Chain Rule. In particular, we use the

Generalized Exponential (base e) Rule with the “inside” function being f (x) • g (x2). To find the derivative of f(:r) • g (:r2), we will use the Product Rule. Then, we will use the Chain Rule again to find the derivative of g (x2):

j'(:r) = ef(x)–g(x2) (d::(f(:r). g (:r2)))

 

(jdx     (x2))) + (x2) (j     (f(x))))

2

= ef(x)-g(x ) (f(x)         (g               g

dx

= ef(x)·g(x2) (J(x) (g’ (:r2)) (;!:(:1:2)) + g (:1:2). J'(x))

= ef(x)·gx( 2) (f (i:) (g’ (i:2)) (2x) + g (i:2) • J’ (i:))

Now, we substitute x = 0:

j’ (0) = ef(O)·g( 02) (f (0) (g’ (02)) (2(0)) + g (02) • f’ (0))

= ef(O)-g(O) (f(0) (g'(0)) (2(0)) + g(0)j'(0))

= ef(O)-g(O) (0 + g(0)j'(0))

=    ef(O)-g(O) (g(0)j’ (0))

Looking at the graphs to obtain the required function values, we see f (0) = 5, g(O) = 1, and

f'(O) = 2 (this is a linear part of the graph so we can calculate the slope). This gives us

j'(0) = ef(O)·g(O) (g(0)f'(0))

= eC5l(1l((1)(2))

= 2e5

Applications

Example 10 The weekly price-demand function of a company that sells :r specialized drill bits for solo objects frozen in carbonite is given by p(x) = 2600e-0•2x, where p(:r) is the price, in dollars, of each drill bit.

(a)   Find p'(lO) and interpret your answer.

(b)   Find the marginal revenue when 10 drill bits are sold each week, and interpret your answer.

Solution:  (a) Let’s start by finding p'(x). Notice this function is of the form e9(x), where g(:r) = -0.2:.r. Using the Chain Rule gives

 

 

 

 

Substituting x = 10 yields

p'(:r) = 2600eg(x)g'(:r)

=  2600e-0•2x(-0.2)

= -520e-0•2x

 

p’ (10) = -520e-0•2C10)

-$70.37 per drill bit

Thus, when 10 drill bits are sold each week, price is decreasing at a rate of $70.37 per drill bit.

(b) To find the marginal revenue when 10 drill bits are sold each week, recall that the marginal revenue function is the derivative of the revenue function. So first, we must find the revenue function:

R(x) = :.r • p(x)

= :r (2600e-0•2x)

= 2600xe-0•2x

 

Thus, R(x) = 2600xe-0•2x gives the revenue, in dollars, when x drill bits are sold each week. Now, to find the derivative R'(J:), we need to use the Product Rule as well as the Chain Rule:

 

R’ (x) = (2600x) (d : (e-0-2x)) + (e-0•2x) ( d! (2600:r))

= 2600x (-0.2e-0•2x) + 2600e-0•2x

= -520J:e-0•2x + 2600e-0•2x

Substituting x = 10 to find the marginal revenue when 10 drill bits are sold gives

R'(lO) = -520(10)e-0•2(10) + 2600e-0•2(10)

-$351.87 per drill bit

 

When 10 drill bits are sold, revenue is decreasing at a rate of $351.87 per drill bit. Recall this answer also gives us an approximation for the revenue earned from selling the next drill bit. In other words, the company will lose approximately $351.87 in revenue when selling the 11th drill bit.

Example 11 The cost, in dollars, of producing :r cameras per week is given by C(:r) = 600 + 100 4:r + 2.

(a)   Estimate the cost when 2-5 cameras are produced.

(b)   Find the exact cost of the 15th camera produced.

(c)   Approximate the cost of the 15th camera produced.

(d)   Calculate and interpret C’ (34).

Solution:  (a) To estimate the cost of producing 25 cameras, we find the total cost of 24 cameras, C(24), and add the approximate cost of producing the 25th camera, which is C'(24).

 

We start by finding C'(x) using the Chain Rule, where we have (g(::r))½ with the “inside” function

g(x) = 4:r + 2:

C'(x) = 0 + 100 ( ( 4:r +2))

dx

= 100( 12(g(::r))-12 • g1(:i:))

=50(4x+2)-½(4)

= 200(4:r + 2)-½

Now, we can calculate the approximation:

0(25) Re; 0(24) + C'(24)

= (600 + 100J4(24) + 2) + 200(4(24) + 2)-½

Re; $1610.15

Hence, the approximate cost of producing 25 cameras is $1610.15.

(b) Remember that to find the exact cost of producing the 15th camera, we do not need to use calculus. We calculate 0(16) – C(15). This difference takes the cost of producing 16 cameras and subtracts the cost of producing the first 15 cameras. This leaves the cost of producing only the 15th camera.

Recalling C(:r) = 600 + 100 4:r + 2, we have

0(16) – 0(15) = (600 + 100J4(16) + 2) – (600 + 100J4(15) + 2)

Re; $25.00

Therefore, the exact cost of producing the 15th camera is $25.00.

(c) To approximate the cost of producing the 15th camera, we substitute one less item into the derivative. Thus, we calculate 0′(15) using the derivative C'(:r) = 200(4:r + 2)-½ we found previously:

C’ (15) = 200(4(15) + 2)-½

Re; $25.40 per camera

Hence, the approximate cost of producing the 15th camera is $25.40. This is close to the exact cost of producing the 15th camera, $25, we found in part b.

(d) C'(34) represents the marginal average cost when 34 cameras are produced. First, we need to find the average cost function, C, and then we can take the derivative to find the marginal average cost function, C’.

We find C(x) first:

C(:r)=   C(:r)

X

600 + 100 4:r + 2

:r

Now, to take the derivative, we start with the Quotient Rule. When finding the derivative of the numerator, we will substitute the derivative of 600 + 100 4:r + 2 we already found in part a:

C'(x)=    x (d  (600 + 100 4x + 2)) – (600 + 100 4:1: + 2) L (x))

;1:2

X ( 200(4:1: + 2)-½) – (600 + 100 4x + 2)

x2 Substituting x = 34 to calculate C’ (34) gives

 

-,            34 ( 200(4(34) + 2)-½) –  (600 + 100)4(34) + 2)

C (34) =                                  (34)2

Re; -$1.03 per camera per camera

 

Thus, when 34 cameras are produced, the average cost per camera is decreasing at a rate of $1.03 per camera.

Example 12 A bank account is opened with a deposit of $50,000. The account has an annual interest rate of 811i1 per year. Find the rate at which the balance of the account is increasing after 6 years if the account is compounded

(a)   continuously.

(b)   semiannually.

Solution: (a) Because we want to find a rate of change, we need to take a derivative. However, we are getting ahead of ourselves; first we need a function! The formula for the balance of an account in which interest is compounded continuously is A = Pert, where P is the principal (or starting) amount, A is the accumulated amount after t years, and r is the interest rate (as a decimal) per year.

With formulas for financial math, we always use rates in their decimal form!

We want a rate of change that is per year, so our independent variable should be t. Letting

P = 50,000 and r = 0.08, we have

 

Now, we can take the derivative using the Chain Rule because we have a function of the form

eg(t), where g(t) = 0.08t :

A'(t) = 50,000e9Ct) • g'(t)

= 50, OOOe0•08t(0.08)

= 4000e0•08t

A'(t) = 4000e0•08t gives us the rate at which the amount in the bank account is increasing after t

years. We need the rate the balance is increasing after 6 years, so we calculate A’ (6):

A'(6) = 4000e0•08<6l

$6464.30 per year

 

After 6 years, the balance of the account is increasing at a rate of $6464.30 per year.

(b) Here, we need to find the rate of change of the balance if the account is compounded semiannually. The formula for an account balance if interest is not compounded continuously is a bit more com­ plicated: A = P (1 + -[;,,)mt, where A, P, r, and t still represent the same quantities as before, and mis the number of compounding periods per year. Because the account compounds semiannually, or two times per year, m = 2 and the function with time as the independent variable is

 

A(t) = 50,000(

2

1 + T0 08)  t

= 50, 000(1 + 0.04)2t

= 50, OOO(l.04)2t

 

Now, we can take the derivative with respect to time using the Chain Rule because the function is of the form 1.049(t), where g(t) = 2t:

A'(t) = 50,000(1.04)9(tl(ln(l.04)) · g'(t)

= 50, 000(1.04)2\ln(l.04))(2)

= 100, OOO(ln(l.04) )(l.04)2t

 

Thus,

 

A’ (6) = 100, OOO(ln(l.04) )(1.04)2(6)

$6279.36 per year

 

After 6 years, the balance of the account is increasing at a rate of $6279.36 per year.

These answers do not give us the amount of money in the accounts after 6 years! These values give us the rate of change of the account balances with respect to time.

Try  It 9

 

 

Alternate Form of the Chain Rule

All of the Generalized Chain Rule formulas discussed in this section are useful when we need to find the derivative of a function that consists of two functions that have already been composed and the result is an “outside” function and an “inside” function. There is another form of the Chain Rule that can be used to find the derivative of a function that consists of functions that have not been composed yet:

Theorem 11 Alternate Form of the Chain Rule

 

Even though most prefer LaGrange notation for derivatives, here we use the Leibnitz notation because we need to pay attention to the variable that we are taking the derivative with respect to. For instance,    tells us the independent variable is u. Let’s look at an example using the Alternate Form of the Chain Rule.

Example 13 If y = 5u4 –  9eu and u = 7 + 12 ln(x), find      using the Alternate Form of the Chain Rule.

Solution: Because we are given two separate functions that have not been composed, using the Alternate Form of the Chain Rule to find the derivative  makes sense! To find , we find the derivative of each function first:

and   –du =0+12(  -1)   = 1-2

 

The Alternate Form of the Chain Rule gives

dy      dy    du

dx      d‘u  dx

dx                 x        x

 

 

 

We want to find ,  and paying attention to the notation tells us that we need to find the derivative of the function y with respect to :r (not u). Thus, we must substitute for ‘ll so the answer is entirely in terms of the variable x. Noting that u = 7 + 12ln(a:), we have

 

Alternatively, we could have used the original form of the Chain Rule in the previous example if we would have composed the functions first. We will demonstrate this technique to show that the two forms of the Chain Rule are equivalent.

Recall that y is a function of u, y = f(u) = 5u4 – 9eu, and 1l is a function of :r,1l = g(x) = 7 + 12 ln(:r). To work the problem this way, we first need to compose the functions to find y = f(g(:r)):

Y = f(g(:r))

= f(7 + 12 ln(x))

= 5(7 + 12 ln(x))4 –  9e7+12111(x)

 

Now, we can use the original form of the Chain Rule to find the derivative. For the first term, we will use the Generalized Power Rule with g(x) = 7 + 12ln(x). For the second term, we will use the Generalized Exponential Rule, where g(:r) also equals 7 + 12ln(x). Writing the function this way, we have

y = 5(7 + 12 ln(:r) )4 – 9e7+12111(x)

= 5(g(:1:) )4 – 9eg(x)

 

Finding the derivative using the corresponding rules gives

y’ = 20(g(x))3 · g'(x) – 9eg(x) • g'(x)

 

Substituting g(:r) = 7 + 12 ln(x) as well as calculating and substituting g'(:r) = 1;,2, we have

 

y’ = 20(7 + 12ln(x))3. ( )     – 9e7+12ln(x). (      )

 

 

X                                                                                l

If we factor 12 from both terms we can write our answer as

 

y’ = (20(7 + 12ln(x))3 – 9e7+121n(x)) • (      )

 

which is identical to the answer we got using the Alternate Form of the Chain Rule in the previous example!

 

These two forms of the Chain Rule are always equivalent.

 

We can also apply the Alternate Form of the Chain Rule regardless of what variables are used as we will see in the next example.

Example 14 If r(k) = ifk, – 1rk and k(a) = log6(a) – 2 • 4a, find  :  using the Alternate Form of the Chain Rule.

 

Solution: To find  :,  we find the derivative of each function first:

 

and

 

The Alternate Form of the Chain Rule gives

dr      dr    dk

da      dk    da

aln(6)

= (!k-2/3 –

3

 

 

 

 

1r) (-1

 

 

 

 

 

–    – 2 (4a ln(4)))

Again, because we want to find 1:, we need the derivative of the function r with respect to a (not k).

6

Thus, we must substitute fork in the answer. Noting that k(a) = log (a) – 2 • 4a, we have

 

–dr= ( 1–;-k-2/3 -1r  ) ( –1 -2(4a  ln(4)))

da        3                    aln(6)

3

=(1 (log6(a) – 2 • 4a)-2/3 – 7r) (aln1(6) – 2 (4a ln(4)))

 

To summarize, we write both forms of the Chain Rule:

 

Remember that as complicated as finding derivatives using the Chain Rule, Product Rule, Quotient Rule, and other rules may be, using these rules is still significantly easier than using the limit definition of the derivative.

 

 

Enrichment: Logarithmic Differentiation

The Generalized Natural Logarithm Rule is useful for finding derivatives of functions with natural logarithms, but it can also be used to find other derivatives. Consider the function y = xx whose graph is shown in Figure 2.5.2.

Figure 2.5.2: Graph of y = :rx

This function has a domain of (0, oo), and its graph has a shape similar to that of a parabola (except with a much more rapid increase). The function has no sharp corners or cusps, no discontinuities, and no vertical

 

tangent lines. Therefore, its derivative should exist everywhere on its domain. However, the function y = :1:x does not fit into any of our previous rules. It turns out that we actually can find a formula for the derivative, but we have to start with something counterintuitive: making the function more complicated! If we let g(x) = ln (:rx), then g(x) = x ln(x), which we can find the derivative of using the Product Rule:

 

g'(x) = x ( )     + (ln(:r))(l)

= 1 + ln(:r)

Alternately, we could have used the Generalized Natural Logarithm Rule to find the derivative of g in its original form, g(x) = ln (xx). Doing so yields

Now, we have created two ways to represent g'(:r):

Setting these two representations equal to each other gives

Because our original goal was to find the derivative of y = xx (i.e., find ix (xx)), we will solve the equation above for ix (xx). Multiplying both sides of the equation by xx gives

In other words,

 

d

dx (:rx) = xx(l + ln(x))

Thus, we have found the derivative of y = :rx!

 

This technique of making something more complicated to access mathematical rules is a common one used in mathematics. In this particular case, it gives rise to a technique called logarithmic differentiation.

 

If we have a function where the independent variable is in both the base and the exponent, such as y = :r:x, we should write the function as the argument of a natural logarithm. Then, we can use the Properties of Logarithms to rewrite the function and take the derivative using previous techniques (i.e., the Product Rule, Quotient Rule, Chain Rule, etc.). Finding the derivative again using the Generalized Natural Logarithm Rule allows us to set the two derivatives equal in order to solve for the desired derivative. We now summarize this process:

Logarithmic Differentiation

 

Let’s see this in action with some examples:

Example 15 Find the derivative of each of the following functions.

(a)    f(x) = 3x5x2-12

(b)

)

f(x) = (l4x 5 + 8x 3 + l 4(x+1)2-15

Solution:  (a) Because the independent variable, :.r:, is in both the base and the exponent off (:r), we will use logarithmic differentiation:

1.  

–

Let g(x) = ln (3:r5x2 12).

2.   Use the Properties of Logarithms to fully expand g(:r):

 

g(x) = ln (3:r5x2-12)

5     12

2= ln(3) +ln (x x –    )

= ln(3) + (5:r2 – 12) ln(:r)

3.   Find g'(x) using previous derivative rules:

 

g'(:r) = d r (ln(3) + (5:r2 – 12) ln(x))

2                                                                      2= 0 + (5:r – 12) ( d Jln(x))) + (ln(x)) (d : (-5x – 12))

5:r2 –  12

=                + (ln(x))(lOx)

:r

4.

;

–

Because g'(x) =  g} by the Generalized Natural Logarithm Rule as well, solving this equa­ tion for f'(x) gives f'(x) = g'(x) · f(x). Thus, we substitute g'(x) = 5×2 12 + (ln(x))(lOx) and f(x) = 3:1:4×2 12 to find f'(x):

f’ (x) = g’ (x) • f (x)

2                                          2

=  (5:i: ;; 12 + (ln(:1:))(lOx)) (:3:1:5x -12)

(b)

= (14x + 8x + 1) (x+1) –    ,

Because the independent variable, x, is in both the base and the exponent of f(x)             5        3           4 2        15

.

we will use logarithmic differentiation:

1.   Let g(x) = ln (( 14:1:5 + Kr3 + 1)

4(x+1)2-15)

2.   Use the Properties of Logarithms to fully expand g(:r):

((

g(x) = ln     14:1:5 + 8x 3 + l)4(x+l)2-15)

=  (4(:i:+1)2-15) (ln(14:r5+8:i:3+1))

3.   Find g'(:1:) using previous derivative rules:

g'(:1:)= (4(:.r:+1)2-15) (d! (ln(14:.r:5+8:i:3+1)))  +(ln(14:.r:5+8:.r:3+1)) (d! (4(:.r:+1)2-15))

5       3

2                A..(14i· +8:r +1)

5        3               (                   (d     )    )

=  (4(x + 1)  –  15) dx        • 5      • •                  +  (ln (14:i: + 8x + 1))     2 · 4(x + 1)    -(x     + 1)    – 0

14:i: + 8×3 + 1                                                              dx

= (4(x + 1)2

70:1.4 + 24:1·2

5        3

5

-15)                   •          + (ln (14:i:

14x + 8x + l

+ 8×3

+ 1)) (8(:1: + 1)(1))

=(4(x+1)2

70:1:4 + 24:.r:2

14:r” + 8:r + 1

5

-15) .•. 3 +(ln(14x

+8:1:3+1))(8:1:+8)

4.   Because g'(x) = 1g] by the Generalized Natural Logarithm Rule as well, solving this equa­ tion for f'(:r) gives J'(:r) = g'(x) · j(J.}

‘                                   ”                                                        14x0+8x3+1

Thus  we substitute q'(x) = (4(x + 1)2 -15) 7o,,;t+24×2  + (ln (14×5 + 8×3 + 1)) (8x + 8) and

2

(

f(x) = •14J.:”C+ 8J.:3+ 1)4(x+1) -15 to find j'(J.}

 

,f (J.:) = ((

4(:r + 1) 2 – 15)    7x0J+.A +2×4+x

 

+( ln( 14x 5

+ 8:r3

 

+ 1))

(8:r + 8))   ((

2

14J.c:” + 8:r3 + 1)4(x+1) -15)

2

14 5       8 3       1

Try It Answers

1.  (a) y’ = 15 (2:r + ex)4 (2 + ex)

(b)

214e- x + x-  (-4:T – 4:C )

2.  (a) y1 =        2 2  2                       3

I      ((ln(x))2) ( e2x-6+sx5(-12x-7+15×4))-e2x-6+sx5(2(1n(x))) ( ¾)

(b) Y =                                    (ln(x))4

3.     (a) y’ = 14ex+log4(x)+4xs (ex+ _xll_n(4) + 32×7)

(b) y’ = (25×3 22) (39:r12        2

13        2

5 3 22(ln(2)) (15:r2))

–

4• a( ) Y –

, _     6x-6×2+1 3×2-2×3+x+12

– 3x

(ln(3))(2x)) + (3x _      3x ) (2 x –

(b) y’ = -48:rll – x!3 –  x_:_3

,     exs-14×2+15x (3×2 – 28x + 15) -16:r 5• (a) Y=                (ex3-14×2+15x _ 8J:2) (ln(4))

(b) y, =

3×2 + 44

–          5:r4 – 3×2 – 1

– l2x3

(x3 + 44:r -18) (ln(7))      (x5 –  x3 – ;r + 1) (ln(7))

6.   y = (6 ln(lO) )x – 9 ln(lO) – 8l

7.     (a) 30e9

(b) 36

8.     (a) 1496

(

b) 14111(37)- ”a?

(ln(37))2

9.     (a) $23.77 per year

(b) $63.74 per year

10.     (a)   i = (17-12(7ln(y)+eY)-4) (t+eY)

(                                                                                                  2

b) !!:Ji.= 14 (5-      (–1L)16)13  (-16(     2x” )15) ((log(x))(12×5)-(2×6)(x1n(l0)))

dx                              log(x)                                    log(x)                        (log(x))

Exercises

 

Basic Skills Practice

 

For Exercises 1 – 3, use the Generalized Power Rule to find the derivative of the function.

1. f(x) = (4x + 12)6                                   2. g(x) = (12×3 + 33x}413

 

For Exercises 4 – 7, use the Generalized Exponential Rule to find  ;  for the function.

4.   y = ex2

5.   y =

r,:;; 13

ev —;;;-

For Exercises 8 – 11, use the Generalized Logarithm Rule to find f’ (:r).

 

8.  J(x)=ln(4+:1:)

9.  f(:r) = ln (9:1:3 -13:r-3/2 + 17)

10.  J(x) = log3 (;½ + 55×2)

11.  f(:r) = log5 (36:r11 -12:r7 + :r3 – 1)

For Exercises 12 – 14, find the slope of the line tangent to the graph off  at the given :1:-value.

12.  J(x) = (7×3 + 55)9 at x = -2

13.  .f (:r) = 5x11) –2 x5 +i 2 x at :r = 0

14.  J(x) = log11 (4:1:2 + 16x) at x = l

For Exercises 15-17, find the equation of the line tangent to the graph of the function at the given :1:-value.

15.  f(:r) = (:r5 –  2) 7 at :r = 1

17.   h(x) = ln (:1:2 + 3x + 3) at x = -l

18.   Given f(x) =} (x3 – 3×2 + 3x)4, find the equation of the line tangent to the graph off at x = 2. Use technology to graph f and the tangent line on the same axes.

For Exercises 19 – 21, find the x-value(s) where the graph off  has a horizontal tangent line.

 

21.   f(x) = ln (:1:2 + llx     + 31)

 

For Exercises 22 – 25, use the Alternate Form of the Chain Rule to find .

22.   y = fo and ·u = 8×3 –  8

23.   y = 8u3 – 8 and u = .Ji

24.   y = ln(z) – 13z and z = :r9 –  3x3 + 1

25.   y = t!i   and t =ex+ 7:1:

26.   The price-demand function for a brand of diploma frame is given by p(:1:) = -(-0.0h – 2)4 + 303, where p(:-r:) is the price, in dollars, per frame when x frames are demanded. At what rate is the price changing when 400 frames are demanded?

27.   The cost of making :1: high end toaster ovens is given by C(:1:) = :1:2 +13x +160 dollars. At what rate is the cost changing when 300 toaster ovens are made. In other words, what is the marginal cost when :1: = 300?

28.   The function s(t) = (t2 )3 represents the position of a particle, in inches, after t seconds. What is the instantaneous velocity of the particle after 1 second?

 

 

 

Intermediate Skills Practice

 

For Exercises 29 – 44, find f'(:i:).

29.   J(x) = (xex + 8x)9

30.

2

J(x) = (x3 -17log (x))8 (2:r + 14)

31.   J(x) = {/(12:1:7 – ex+ 7!-)4. gx

32. J(:1:) = g::1 1;

33.   J(x) = (‘n(:t 7×4 )6

 

 

37. J(x) = 7x/(x–7)

1       x2–1

38- f(:r:) = (17 +17)

 

39.       f (:1:) = (i:4el2x-3 – 9)-1

 

40.      J(x) = (8×2) (ln (:1:2 + 49))

41.      J(x) = (ex – 1) (log(9 – fa))

42.

=

f(     )              3

34.   J(:1:) =

3

6

3 1x +7x       3

(4×5+x)

5x -125x

X           ln(7x+l4)

44.

_  ln(6×2+13x-2)

f(X )-            (x3-l)e

For Exercises 45 – 47, (a) use the Properties of Logarithms to fully expand f(:1:) and (b) find f'(:1:).

4-5. J(x) = ln ((x + 4)7(x -10)14)

46-

l      ( (5x-77)0)

f( :1:) =

og5      (18x+9)3

 

For Exercises 48 – 50, find the slope of the line tangent to the graph off at the given :1:-value.

48. f(x) = (:1: + 10)(3:1: + 22)8 at x = -7

49. J(:1:) = eCx-9)’7 at :1: = 10

 

,50. J(x) = In( = !j'”) at :1: = 0

 

51. Given f(:1:) = ln (5×2 + 12x + 3), find the equation of the line tangent to the graph off at x = 0. Use technology to graph .f and the tangent line on the same axes.

For Exercises 52 and 53, find the :1:-value(s) where the graph off has a horizontal tangent line.

53.   .f(.r) = ln ( sx2-4 +1)

 

For Exercises 54 and 55, find the :r-value(s) where the line tangent to the graph off has the indicated slope.

54.    f(x) = ln (x2 – 5x – -5); slope is 1                      -55. f(x) = 27x + e2x3+33×2+15sx; slope is 27

 

ForExercises56-59,find ;.

56.   y =2u + 1 and u = 4x9

57.   y = ,5u – 1:3u3 and 1l = ln (:1:2 + 36)

58.   y = (w2 + 1) ew and w = 2:r3 + 5

59.   y = 3t2-2t+1 andt = 2×2 + 32 60. The daily profit function for the food truck Crepes by Monica is given by the function

P(:r) = 10 (x2 –  1000)113 –  287

dollars when x crepes are sold. Find P'(33) and interpret your answer.

60.   The weekly cost function for Susan’s Sticky Notes is given by

C(x) = 20 ln (3×2 + lOx + 25)

dollars, where :r is the number of boxes of sticky notes produced.

(a)   Find the marginal cost when 160 boxes of sticky notes are produced, and interpret your answer.

(b)   Approximate the cost of producing the 200th box of sticky notes.

(c) Find the exact cost of producing the 175th box of sticky notes.

(d)    Estimate the cost if 130 boxes of sticky notes are produced.

(e)   Find the exact cost if 250 boxes of sticky notes are produced.

61.   A corporation deposits $80, 000 in a new bank account that earns interest at an annual rate of 2.7% per year and compounds continuously. The formula for the amount in the account after t years is given by

A(t) = 80, 000eo.o21t

dollars. At what rate is the balance of the account growing after 7 years?

62.   Mahesh is saving to take a vacation. He deposited $1200 in a bank account that earns interest at an annual rate of 3% per year and compounds monthly. The formula for the amount in the account after t years is given by

A(t) = 1200(1.0025)12t

dollars. How fast is Mahesh’s account balance growing after 2 years?

63.   A pteranodon is flying by. The scientist observing it notices that its position, in feet, from her after t

seconds is given by the function

s(t) = 30(8t + 3)213 + 10

At what speed is the pteranodon flying after 30 seconds? Note: The speed is given by the absolute value of the rate of change of s(t). Round your answer to two decimal places, if necessary.

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

(a)    h'(-1), where h(:r) = g(J(x))

(b)   j'(-5), where j(x) = 3x – f(f(x))

(c)   k'(2), where k(:r) = g (:r3)

 

65.   The table below shows values of f(;r), g(x), f'(x), and g'(x) for certain values of x. Use the information in the table to find each of the following.

 

X	f(x)	g(x)	f’ (:r)	g'(x)
-2	1	1	-2	-2
-1	2	1	-1	1
0	2	2	-1	-1
1	1	-2	0	-2
2	0	-2	-1	1

 

(a)   h'(0), where h(x) = g(f(x))

(b)   j'(-2), where j(x) = f(:( 3)

(c)   k'(-1), where k(x) = 3g (:r3-)

 

 

 

ln(f(:r))

 

 

Mastery Practice

For Exercises 67 – 77, find *.

 

_        (  (14x – 7x

2 8

3

) 2x

 

-1)

67.  y   – log2         (x–   33)7. 24x+5

 

 

68.        y =

( ll:r5+ 54×2-l6)7

xln(x)

8u3                                3 2

69.        y =                and ‘U = (4:i: – 12) 1

ln(u+l)

70.        y = {I4:r3 –  3:r-4

=

9e7-v’x4

71.       y     25 x2-3x+4

72.        y = (8 – x3) 43×2-64

73.        y = 74t17 + 35t7 – 4lt + 2 and t = (14ln(:r) + *)3

 

74.     = In

y          ( ijx2

1

+ 16 • (3X  + 19)

10 )

75.   y=  e2x11+44

76.   y = J(Jr:i:-2) (log4 (9J:2 –  5ex))

77.

eu , and

y = v’t+S, t =    2        u = ln (x2 + 1)

78.

5 x – at

Find the equation of the line tangent to the graph of f(x) = 2 2 1          x = l. Use technology to graph

f and the tangent line on the same axes.

 

For Exercises 79 – 81, find the :1:-value(s) where the graph off has a horizontal tangent line.

 

79.  J(x)=  (x + 2)1139

(:r – 2)

80.   f(x) = log8 (3×4 –  8×3 –  18:r2 + 140) + 14

81.   f(:r) = x2esx2+100

For Exercises 82 and 83, find the :r-value(s) where the graph off  has the indicated slope.

82.

ln(lO)

f(x) = log (x2 + 15:r + 15,) • slope is -1-

83.

–

f(x) = 23 + 13x – 12×3 4sx; slope is 13

84.   Find the value(s) of x where the function f (;r) = 5 (x3 – 8) 38 -18x has an instantaneous rate of change of -18.

85.   A dust particle is D(t) = 13 –  (0.015t3 –  0.75t)413 inches above the ground after t seconds. How fast is the particle moving after 6 seconds? Round your answer to three decimal places, if necessary.

86.   The monthly cost function for vibranium mining operations in the small nation of Wakanda is given by

C(t) = log9 (0.06t4 + 23t2 + 30)

thousands of Wakandan dollars when the mines have been open fort hours this month. Find C'(50) and interpret your answer. Round to three decimal places, if necessary.

87.   Pack Men is a moving company specializing in safely transporting arcade game cabinets. The yearly profit is P(:r) = 2.2J x7/3 + 90 – 3.4:r – 200 hundreds of dollars when :r moves are completed. Find the

marginal profit when 3299 moves are completed, and interpret your answer. Round to four decimal places, if necessary.

88.   The price-demand function for a brand of diploma frame is given by p(:r) = -(-0.0h – 2)4 + 303, where p(:1:) is the price, in dollars, per frame when x frames are demanded.

(a)   Find R'(150), where R(x) is the revenue, in dollars, when :1: frames are demanded, and interpret your answer.

(b)   Find the marginal average revenue when 150 frames are sold, and interpret your answer.

89.   Generic Corp, a manufacturer of doodads, has a cost function given by C(x) = (0.02×2 + 0.18:1: + 30.28) 719

dollars when x doodads are made.

(a)   Find the rate of change of cost when 116 doodads are made, and interpret your answer.

(b)   Approximate the cost of making the 116th doodad.

(c) Find the exact cost of making the 116th doodad.

(d) Estimate the cost if 116 doodads are made.

(e) Find the exact cost if 116 doodads are made.

90.   A bank account has an initial balance of $400. The account earns interest at an annual rate of 3.24% per year compounded continuously. How fast is the balance of the account growing after 7 years?

91.   Walton Martinez is planning on opening a store and is saving money by placing $20, 000 in a bank account that compounds quarterly and has an annual interest rate of 4.2% per year. At what rate is his account balance growing after 3 years?

92.   The table below shows values of f(:1:), g(x), f'(x), and g'(:1:) for certain values of :1:. Use the information in the table to find each of the following.

 

:r	f(:r)	g(x)	f'(:r)	g’ (:r)
-10	5	5	-8	-5
-5	10	5	-4	DNE
0	10	10	-3	2
5	5	-10	0	-2
10	0	1	-8	3