Jacques I. Mathematics for Economics and Business

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9 (a) Show that, at a stationary point of an average cost function, average cost equals marginal cost.

(b) Show that if the marginal cost curve cuts the average cost curve from below then average cost is

a minimum.

10 In a competitive market the equilibrium price, P, and quantity, Q, are found by setting Q

= Q

in the supply and demand equations

P = aQ

+ b (a > 0, b > 0)

P =−cQ

+ d (c > 0, d > 0)

If the government levies an excise tax, t, per unit, show that

Q =

Deduce that the government’s tax revenue, T = tQ, is maximized by taking

t =

d − b

d − b − t

a + c

Differentiation

330

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section 4.8

The derivative of the

exponential and natural

logarithm functions

In this section we investigate the derived functions associated with the exponential and

natural logarithm functions, e

and ln x. The approach that we adopt is similar to that used

in Section 4.1. The derivative of a function determines the slope of the graph of a function.

Consequently, to discover how to differentiate an unfamiliar function we ﬁrst produce an

accurate sketch and then measure the slopes of the tangents at selected points.

Objectives

At the end of this section you should be able to:

 Differentiate the exponential function.

 Differentiate the natural logarithm function.

 Use the chain, product and quotient rules to differentiate combinations of these

functions.

 Appreciate the use of the exponential function in economic modelling.

Advice

The functions, e

and ln x were first introduced in Section 2.4. You might find it useful to

remind yourself how these functions are defined before working through the rest of the

current section.

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Differentiation

332

Example

Complete the following table of function values and hence sketch a graph of f(x) = e

x −2.0 −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5

f(x)

Draw tangents to the graph at x =−1, 0 and 1. Hence estimate the values of f ′(−1), f ′(0) and f ′(1). Suggest

a general formula for the derived function f ′(x).

Solution

Using a calculator we obtain

x −2.0 −1.5 −1.0 −0.5 0.0 0.5 1.0 1.5

f(x) 0.14 0.22 0.37 0.61 1.00 1.65 2.72 4.48

The corresponding graph of the exponential function is sketched in Figure 4.31. From the graph we see

that the slopes of the tangents are

Figure 4.31

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f ′(−1) ==0.4

f ′(0) ==1.0

f ′(1) ==2.7

These results are obtained by measurement and so are quoted to only 1 decimal place. We cannot really

expect to achieve any greater accuracy using this approach.

The values of x, f(x) and f ′(x) are summarized in the following table. The values of f (x) are rounded to

1 decimal place in order to compare with the graphical estimates of f ′(x).

x −10 1

f(x) 0.4 1.0 2.7

f ′(x) 0.4 1.0 2.7

Notice that the values of f(x) and f ′(x) are identical to within the accuracy quoted.

These results suggest that the slope of the graph at each point is the same as the function value at that

point: that is, e

differentiates to itself. Symbolically,

if f (x) = e

then f ′(x) = e

or, equivalently,

if y = e

then = e

1.35

0.50

0.20

0.50

4.8 • The derivative of the exponential and natural logarithm functions

333

Practice Problem

1 Use your calculator to complete the following table of function values and hence sketch an accurate

graph of f(x) = ln x.

x 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00

f(x) 0.41 1.25

Draw the tangents to the graph at x = 1, 2 and 3. Hence estimate the values of f′(1), f′(2) and f ′(3).

Suggest a general formula for the derived function f′(x).

[Hint: for the last part you may find it helpful to rewrite your estimates of f′(x) as simple fractions.]

In fact, it is possible to prove that, for any value of the constant m,

if y = e

then = me

and

if y = ln mx then =

In particular, we see by setting m = 1 that

differentiates to e

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and that

ln x differentiates to

which agree with our practical investigations.

Differentiation

334

Example

Differentiate

(a) y = e

(b) y = e

−7x

Solution

(a) Setting m = 2 in the general formula shows that

if y = e

then = 2e

Notice that when exponential functions are differentiated the power itself does not change. All that

happens is that the coefﬁcient of x comes down to the front.

(b) Setting m =−7 in the general formula shows that

if y = e

−7x

then =−7e

−7x

if y = ln 5x then =

Notice the restriction x > 0 stated in the question. This is needed to ensure that we do not attempt to

take the logarithm of a negative number, which is impossible.

(d) Setting m = 559 in the general formula shows that

if y = ln 559x then =

Notice that we get the same answer as part (c). The derivative of the natural logarithm function does not

depend on the coefﬁcient of x. This fact may seem rather strange but it is easily accounted for. The third rule

of logarithms shows that ln 559x is the same as

ln 559 + ln x

The ﬁrst term is merely a constant, so differentiates to zero, and the second term differentiates to 1/x.

Practice Problem

2 Differentiate

(a)

y = e

(b) y = e

−x

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The chain rule can be used to explain what happens to the m when differentiating e

. The

outer function is the exponential, which differentiates to itself, and the inner function is mx,

which differentiates to m. Hence, by the chain rule,

if y = e

then = e

× m = me

Similarly, noting that the natural logarithm function differentiates to the reciprocal function,

if y = ln mx then =×m =

The chain, product and quotient rules can be used to differentiate more complicated func-

tions involving e

and ln x.

4.8 • The derivative of the exponential and natural logarithm functions

335

Example

Differentiate

(a) y = x

(b) y = ln(x

+ 2x + 1) (c) y =

Solution

(a) The function x

involves the product of two simpler functions, x

and e

, so we need to use the pro-

duct rule to differentiate it. Putting

u = x

and v = e

gives

= 3x

and = 2e

By the product rule

= u + v

= x

[2e

] + e

[3x

]

= 2x

+ 3x

There is a common factor of x

, which goes into the ﬁrst term 2x times and into the second term

3 times. Hence

= x

(2x + 3)

(b) The expression ln(x

+ 2x + 1) can be regarded as a function of a function, so we can use the chain rule

to differentiate it. We ﬁrst differentiate the outer log function to get

and then multiply by the derivative of the inner function, x

+ 2x + 1, which is 2x + 2. Hence

2x + 2

+ 2x + 1

+ 2



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is the quotient of the simpler functions

u = e

and v = x

+ 2

for which

= 3e

and = 2x

By the quotient rule

(3x

− 2x + 6)

+ 2)

[3(x

+ 2) − 2x]

+ 2)

+ 2)(3e

) − e

(2x)

+ 2)

− u

dx dx

+ 2

Differentiation

336

Practice Problem

3 Differentiate

(a)

y = x

ln x

(b) y = e

ln x

x + 2

Advice

If you ever need to differentiate a function of the form:

ln(an inner function involving products, quotients or powers of x)

then it is usually quicker to use the rules of logs to expand the expression before you

begin. The three rules are

Rule 1 ln(x × y) = ln x + ln y

Rule 2 ln(x ÷ y) = ln x − ln y

Rule 3 ln x

= m ln x

The following example shows how to apply this ‘trick’ in practice.

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Exponential and natural logarithm functions provide good mathematical models in many

areas of economics and we conclude this chapter with some illustrative examples.

4.8 • The derivative of the exponential and natural logarithm functions

337

Example

Differentiate

(a) y = ln(x(x + l)

) (b) y = ln

Solution

(a) From rule 1

ln(x(x + 1)

) = ln x + ln(x + 1)

which can be simpliﬁed further using rule 3 to give

y = ln x + 4 ln(x + 1)

Differentiation of this new expression is trivial. We see immediately that

If desired the ﬁnal answer can be put over a common denominator

+= =

(b) The quickest way to differentiate

y = ln

is to expand ﬁrst to get

y = ln x − ln(x + 5)

1/ 2

(rule 2)

= ln x −

/2 ln(x + 5) (rule 3)

Again this expression is easy to differentiate:

=−

If desired, this can be written as a single fraction:

−= =

x + 10

2x(x + 5)

2(x + 5) − x

2x(x + 5)

2(x + 5)

(x + 5)

5x + 1

x(x + 1)

(x + 1) + 4x

x(x + 1)

x + 1

(x + 5)

Practice Problem

4 Differentiate the following functions by first expanding each expression using the rules of logs:

(a)

y = ln(x

(x + 2)

) (b) y = ln

2x + 3

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Differentiation

338

Example

A ﬁrm’s short-run production function is given by

Q = L

−0.01L

Find the value of L which maximizes the average product of labour.

Solution

The average product of labour is given by

== =Le

−0.01L

To maximize this function we adopt the strategy described in Section 4.6.

Step 1

At a stationary point

= 0

To differentiate Le

−0.01L

, we use the product rule. If

u = L and v = e

−0.01L

then

= 1 and =−0.01e

−0.01L

By the product rule

= u + v

= L(−0.01e

−0.01L

) + e

−0.01L

= (1 − 0.01L)e

−0.01L

We know that a negative exponential is never equal to zero. (Although e

−0.01L

gets ever closer to zero as L

increases, it never actually reaches it for ﬁnite values of L.) Hence the only way that

(1 − 0.01L)e

−0.01L

can equal zero is when

1 − 0.01L = 0

which has solution L = 100.

Step 2

To show that this is a maximum we need to differentiate a second time. To do this we apply the product

rule to

(1 − 0.01L)e

−0.01L

taking

u = 1 − 0.01L and v = e

−0.01L

d(AP

)

d(AP

)

−0.01L

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for which

=−0.01 and =−0.01e

−0.01L

Hence

= u + v

= (1 − 0.01L)(−0.01e

−0.01L

) + e

−0.01L

(−0.01) = (−0.02 + 0.0001L)e

−0.01L

Finally, putting L = 100 into this gives

=−0.0037

The fact that this is negative shows that the stationary point, L = 100, is indeed a maximum.

(AP

)

(AP

)

4.8 • The derivative of the exponential and natural logarithm functions

339

Practice Problem

5 The demand function of a good is given by

Q = 1000e

−0.2P

If fixed costs are 100 and the variable costs are 2 per unit, show that the profit function is given by

π=1000Pe

−0.2P

− 2000e

−0.2P

− 100

Find the price needed to maximize profit.

Example

A ﬁrm estimates that the total revenue received from the sale of Q goods is given by

TR = ln(1 + 1000Q

)

Calculate the marginal revenue when Q = 10.

Solution

The marginal revenue function is obtained by differentiating the total revenue function. To differentiate

ln(1 + 1000Q

) we use the chain rule. We ﬁrst differentiate the outer log function to get

and then multiply by the derivative of the inner function, 1 + 1000Q

, to get 2000Q. Hence

MR ==

At Q = 10,

MR ==0.2

2000(10)

1 + 1000(10)

2000Q

1 + 1000Q

d(TR)

1 + 1000Q

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