Sums and Differences of Functions
[ad_1]Introduction
Economists are interested in the effects of change. Therefore, the concept of
the derivative of a function, which provides information about how a function
changes in response to changes in the independent variable, is an important
one in economic analysis. For example, the derivative of a production function
provides information about the manner in which the output of a production
process changes as the number of workers employed by the company changes.
Differentiation is the mathematical tool that allows us to quantify such rates of
change. As we will see in Chapter 7, differentiation is also an important tool in
the determination of the maximum or minimum values of economic functions
such as profit and cost.
In Chapter 2, some linear functions in economics such as linear demand
and supply functions were introduced. These functions are characterized by
their graphs being lines having a constant slope, i.e., the gradient is constant
irrespective of the value of the independent variable. We say that the rate
of change of the function is independent of the point where it is measured.
Furthermore, the slope or gradient of a linear function may be determined
by taking any two points on the straight line and calculating the ratio of the
change in the vertical direction with respect to the change in the horizontal
direction as the value of the independent variable increases. The corresponding
situation for a nonlinear function is quite different, however, and the rate of
change of a nonlinear function varies as one moves along the curve given by its
109
110 Elements of Mathematics for Economics and Finance
P
Q
Δx
tangent at P
R
y=f(x)
Δy
Figure 6.1 The graph of a nonlinear function in which the tangent at the
point P is drawn.
graph.
In Fig. 6.1, we show part of the graph of a nonlinear function y = f(x).
On this graph, we have drawn the tangent to the curve at the point P. The
tangent to a curve at a point P is the straight line that passes through P and
that just touches the curve at this point. The slope or gradient of a curve
y = f(x) at P is then defined to be the gradient of the tangent to the curve
at P. It is a measure of the prevailing rate of change of y relative to x at P.
We can see from Fig. 6.1 that the gradient of a nonlinear function varies as we
move along the curve.
In mathematics, we use the notation f(a) (pronounced f primed of a) to
represent the slope of the tangent to the function f at x = a. The slope of the
tangent to a function is called the derivative of the function – corresponding
to each value of x there is a uniquely defined derivative f(x). Therefore, the
derivative of a function of x is also a function of x.
If y = f(x), then an alternative notation for the derivative of a function is
dy
dx
.
This is pronounced ‘dee y by dee x’. Note that this is a single entity not to
- Differentiation 111
be manipulated in any sense and represents the derivative of y with respect to
x. If, for example, f(x) = x2, then it is natural to use f(x) to represent the
derivative of f(x), whereas if y = x2 is used then dy/dx is more appropriate.
Consider the function y = f(x). The graph of this function is shown in
Fig. 6.1. The slope or gradient of the function at the point P : (x, f(x)) is
the slope of the tangent to the graph of the function at P (see Fig. 6.1). This
slope can be approximated by the slope of the chord PQ where Q is the point
(x + Δx, f(x + Δx)). (A chord is a straight line joining any two points on a
curve.) So the horizontal distance from P to Q is Δx. If x is a variable, the
notation Δx will denote a small change in x. Therefore,
the slope of PQ = QR
PR
= f(x + Δx) − f(x)
(x + Δx) − x
= f(x + Δx) − f(x)
Δx
If Q is allowed to approach P in which case Δx approaches 0, the slope of the
chord PQ approaches the slope of the tangent at P, i.e.,
Slope of the Tangent at P = lim
Δx→0
f(x + Δx) − f(x)
Δx
. (6.1)
The value of this limit, if it exists, is known as the derivative of the function
f at x and is written f(x) or dy/dx. Thus, we have
f
(x) = lim
Δx→0
f(x + Δx) − f(x)
Δx
. (6.2)
So the derivative of a function at a point is the ratio of the change in y to the
change in x between the point and a point that it infinitesimally close to it. So
the derivative measures the instantaneous rate of change of the function.
If Δy denotes the change y corresponding to the change Δx in x, then
Δy = f(x + Δx) − f(x),
with f(x + Δx) being the value of y = f(x) when the value of x changes to
x + Δx. Therefore,
f
(x)
or dy
dx
= lim
Δx→0
Δy
Δx
. (6.3)
Therefore, for a small change Δx in x and corresponding small change Δy in
y, we have that
dy
dx
≈ Δy
Δx
,
112 Elements of Mathematics for Economics and Finance
or
Δy ≈ Δx
dy
dx
. (6.4)
This makes sense if dy/dx is regarded as the rate of change of y relative to x. In
particular, dy/dx can be regarded as approximately the change in y resulting
from a 1 unit increase in x (provided the value of x is relatively large so that
1 unit is relatively small). The approximation (6.4) is known as the small
increments formula.
The process of finding the derivative of a function is known as differentiation.
The definition of a function may be used to determine the derivative of
a given function. This process is known as differentiation from first principles.
For example, if f(x) = x2, then using (6.2) we have
f
(x) = lim
Δx→0
(x + Δx)2 − x2
Δx
= lim
Δx→0
(x2 + 2xΔx + (Δx)2) − x2
Δx
= lim
Δx→0
2xΔx + (Δx)2
Δx
= lim
Δx→0
2xΔx
Δx
+
(Δx)2
Δx
= lim
Δx→0
(2x + Δx)
= 2x.
The process of determining the derivative of a function from first principles
can be quite time consuming and involve lengthy mathematical calculations.
Fortunately, there is a more rapid route to determining the derivative of the
sorts of functions that we encounter in economics based on a number of rules,
known as rules of differentiation. Some of these rules will be derived in the next
section using the definition of the derivative of a function (6.2), but others will
be stated simply without justification.
Example 6.1
Differentiate y = f(x) = x2 and use the small increments formula to estimate
the change in y if x changes from 1 to 1.01. Calculate also the actual change
in y. - Differentiation 113
Solution. We have already shown that dy/dx = f(x) = 2x and so f(1) = 2.
If x increases from 1 to 1.01, then Δx = 0.01. Therefore, we can estimate the
change in y using the small increments formula (6.4) as
Δy ≈ Δx × f
(1) = 0.01 × 2 = 0.02.
The actual change Δy in y is f(1.01) − f(1) = 1.0201 − 1 = 0.0201.
6.2 Rules of Differentiation
In this section, we show how to differentiate functions without having to use
the definition (6.2). A few rules are sufficient to differentiate all the functions
encountered in this book.
6.2.1 Constant Functions
Consider the constant function f(x) = k, where k is a constant. Using the
definition of a derivative (6.2), we have
f
(x) = lim
Δx→0
f(x + Δx) − f(x)
Δx
= lim
Δx→0
k − k
Δx
= lim
Δx→0
0
Δx
= 0.
Thus, if f(x) = k then f(x) = 0. For example, if f(x) = 8, then f(x) = 0.
This rule is obvious if f(x) is regarded as the rate of change of f(x) relative
to x. In this case, f(x) is constant.
114 Elements of Mathematics for Economics and Finance
6.2.2 Linear Functions
Consider the linear function f(x) = ax+b, where a and b are constants. Using
the definition of a derivative (6.2), we have
f
(x) = lim
Δx→0
f(x + Δx) − f(x)
Δx
= lim
Δx→0
(a(x + Δx) + b) − (ax + b)
Δx
= lim
Δx→0
aΔx
Δx
= lim
Δx→0
a
= a.
Thus, if f(x) = ax + b then f(x) = a. This is the linear function rule. For
example, if f(x) = 3x+2, then f(x) = 3, and if f(x) = 5−1
4x, then f(x) = −1
4 .
6.2.3 Power Functions
Consider the power function f(x) = kxn, where k is a constant and n is any
real number. The derivative of this power function is given by f(x) = knxn−1.
So to obtain the derivative of a power function, we multiply it by the power
and reduce the original power by one. For example, if f(x) = 4×3, then f(x) =
4×3×x3−1 = 12×2, and if f(x) = x4, then f(x) = 4×4−1 = 4×3. When k = 1,
an important special case of this rule is realized, i.e., if f(x) = xn, then
f
(x) = nxn−1. (6.5)
This rule, known as the power function rule, is derived using the definition
(6.2). Since it involves the expansion of (x + Δx)n and some algebra, we omit
the details here.
6.2.4 Sums and Differences of Functions
The rules we have introduced so far can be used to generate the derivatives of
polynomials, the terms of which are power functions. Consider the function f,
which is the sum of two functions g and h, i.e., f(x) = g(x) + h(x). Using the - Differentiation 115
definition of a derivative (6.2) we have
f
(x) = lim
Δx→0
f(x + Δx) − f(x)
Δx
= lim
Δx→0
(g(x + Δx) + h(x + Δx) − (g(x) + h(x))
Δx
= lim
Δx→0
(g(x + Δx) − g(x)) + (h(x + Δx) − h(x))
Δx
= lim
Δx→0
(g(x + Δx) − g(x))
Δx
- lim
Δx→0
(h(x + Δx) − h(x))
Δx
= g
(x) + h
(x).
Thus, if f(x) = g(x) + h(x), then
f
(x) = g
(x) + h
(x).
This is intuitively clear when derivatives are viewed as rates of change: the rate
of change relative to x of two functions of x is the sum of their rates of change.
Similarly, we can show that if f is the difference of two functions g and h, i.e.,
f(x) = g(x) − h(x), then
f
(x) = g
(x) − h
(x).
Thus, the derivative of a sum of two functions is equal to the sum of the
derivatives of the individual functions. Similarly, the derivative of the difference
of two functions is equal to the difference of the derivatives of the two functions.
For example, if f(x) = 12×5 − 4×4, then f(x) = 60×4 − 16×3, and if f(x) =
9×2 + 2x − 3, then f(x) = 18x + 2.
Example 6.2
Differentiate each of the following functions:
- f(x) = 9x − 6,
- y = −9x−4,
- f(x) = x8 + 8×6 + 11.
116 Elements of Mathematics for Economics and Finance
Solution. - This is a linear function (see Section 6.2.2) with a = 9 and b = −6. Therefore,
using the linear function rule, we have f(x) = 9. - This is a power function (see Section 6.2.3) with k = −9 and n = −4.
Therefore, using the power function rule, we have
dy
dx
= (−9)(−4)x
−4−1 = 36x
−5 =
36
x5 . - This is an example of a polynomial function comprising two power functions
and a constant function. Therefore, using the rule for the sum of functions
in conjunction with the power function and constant function (see Section
6.2.1) rules, we have
f
(x) = 8×8−1 + 8 × 6×6−1 + 0 = 8×7 + 48×5.
Note that the linear function rule can be deduced from a combination of
the rule for the differentiation of the sum of two functions and the constant
function and power function rules.
6.2.5 Product of Functions
Suppose that y = uv where u and v are functions of x. Let Δu, Δv, and Δy
denote very small changes in u, v, and y, respectively, that correspond to a
small change Δx in x. Then
y + Δy = (u + Δu)(v + Δv)
= uv + uΔv + vΔu + ΔuΔv.
Since y = uv,
Δy = uΔv + vΔu + ΔuΔv.
We can ignore the term ΔuΔv since it is the product of two very small changes
and therefore negligible. Therefore,
Δy
Δx
= u
Δv
Δx
- v
Δu
Δx - Δu
Δx
Δv.
As Δx → 0,
Δy
Δx
→ dy
dx
,
and
u
Δv
Δx - v
Δu
Δx - Δu
Δx
Δv → u
dv
dx - v
du
dx
,
- Differentiation 117
since (Δu/Δx)Δv → 0 as Δx → 0. Thus, we obtain the product rule for
differentiation: if y = uv, then
dy
dx
= u
dv
dx
- v
du
dx
. (6.6)
6.2.6 Quotient of Functions
Suppose that y = u/v where u and v are functions of x. Let Δu, Δv, and Δy
denote small changes in u, v, and y, respectively, that correspond to a very
small change Δx in x. Thus Δu → 0 and Δv → 0 as Δx → 0. Then
y + Δy = u + Δu
v + Δv
.
Subtracting y = u/v from both sides of this equation yields
Δy = u + Δu
v + Δv
− u
v
.
Simplifying the fraction on the right-hand side of this equation gives
Δy = v(u + Δu) − u(v + Δv)
v(v + Δv)
= vΔu − uΔv
v(v + Δv) .
Therefore,
Δy
Δx
v Δu
Δx
− uΔv
Δx
v(v + Δv) .
Finally, letting Δx → 0, we obtain the quotient rule for differentiation: if
y = u/v then
dy
dx
v du
dx
− udv
dx
v2 . (6.7)
6.2.7 The Chain Rule
Suppose that y is a function of u, i.e., y = f(u), and that u in turn is a function
of x, i.e., u = g(x). We say that y is a function of a function and to express y
as a function of x we write
y = f(g(x)).
If Δy and Δu denote changes in y and u, respectively, that correspond to a
small change Δx in x, then
Δy
Δx
= Δy
Δu
Δu
Δx
.
118 Elements of Mathematics for Economics and Finance
Then, as Δx → 0, we obtain the so-called chain rule:
dy
dx
= dy
du
du
dx
= f
(u)g
(x). (6.8)
We may also write this in terms of derivatives of f and g and then express the
result solely in terms of a function of x, i.e.,
dy
dx
= f
(g(x))g
(x). (6.9)
As an illustration of the use of the chain rule to obtain the derivative of a
function, consider y = (x2 + 3x + 2)5. If we let u = x2 + 3x + 2, then y = u5.
Differentiating u with respect to x and y with respect to u, we obtain
dy
du
= 5u4,
du
dx
= 2x + 3.
Then using the chain rule yields
dy
dx
= dy
du
du
dx
= 5u4(2x + 3)
= 5(x2 + 3x + 2)4(2x + 3).
If we put x = y in (6.8), we obtain
dy
dy
= 1 = dy
du
du
dy
.
It follows that
du
dy
1
dy
du
. (6.10)
Example 6.3
Find the derivative of each of the following functions:
- f(x) = (2×3 + 1)(x2 − 3x) and evaluate f(1),
- f(x) =
5×2 + 3
x2 + 1
and evaluate f(0), - y = (7×4 + 2)6 and evaluate dy/dx when x = 0.
- Differentiation 119
Solution. - To differentiate this function, we use the product rule with u = 2×3 + 1
and v = x2 − 3x. Now du/dx = 6×2 and dv/dx = 2x − 3. Therefore, using
the product rule (6.6), we have
f
(x) = (2×3 + 1)(2x − 3) + 6×2(x2 − 3x),
which, after some simplification, gives
f
(x) = 10×4 − 24×3 + 2x − 3.
Finally, f(1) = 10 − 24 + 2 −3 = −15. - To differentiate this function, we use the quotient rule with u = 5×2 + 3
and v = x2 + 1. Now du/dx = 10x and dv/dx = 2x. Therefore, using the
quotient rule (6.7), we have
f
(x) =
(10x)(x2 + 1) − (5×2 + 3)(2x)
(x2 + 1)2
10×3 + 10x − 10×3 − 6x
(x2 + 1)2
4x
(x2 + 1)2 .
Evaluating this derivative when x = 0 gives f(0) = 0.
- To differentiate y = (7×4+2)6, we use the chain rule (6.8). Let u = 7×4+2
then y = u6. Now
dy
du
= 6u5,
du
dx
= 28×3.
Therefore,
dy
dx
= (6u5)(28×3) = 168(7×4 + 2)5×3.
When x = 0, dy/dx = 0.
6.3 Exponential and Logarithmic Functions
Let f be the exponential function f(x) = eg(x), where g(x) is some function of
x. Then the derivative of f is
f
(x) = g
(x) eg(x). (6.11)
For example, if f(x) = ex2, then f(x) = 2xex2 since g(x) = x2 and g(x) =
2x. When g(x) = 1, we have the important result that the derivative of the
120 Elements of Mathematics for Economics and Finance
exponential function ex is itself, i.e., ex, since g(x) = 1. More generally, if
f(x) = ekx, where k is a constant, then
f
(x) = kekx.
For example, if f(x) = e−2x, then f(x) = −2e−2x.
Let f be the natural logarithmic function f(x) = lng(x), then the derivative
of f is
f
(x) = g(x)
g(x) . (6.12)
For example, if f(x) = ln 6×2 then
f
(x) =
12x
6×2 =
2
x
since g(x) = 6×2 and g(x) = 12x. When g(x) = x we have the result that the
derivative of ln x is 1/x since g(x) = 1.
We display these rules in Table 6.1.
Example 6.4
Find the derivative of each of the following functions: - f(x) = 3e7−2x,
- f(x) = ln(x2 + 6x + 2).
Solution. - If f(x) = 3e7−2x, then g(x) = 7 − 2x. Since g(x) = −2, then
f
(x) = 3g
(x)eg(x) = −6e7−2x.
Table 6.1 Derivatives of the exponential and logarithmic functions.
f(x) f(x)
eg(x) g(x)eg(x)
ex ex
ln g(x) g(x)
g(x)
ln x
1
x - Differentiation 121
- If f(x) = ln(x2 + 6x + 2), then g(x) = x2 + 6x + 2. Since g(x) = 2x + 6,
then
f
(x) = g(x)
g(x)
2x + 6
x2 + 6x + 2.
6.4 Marginal Functions in Economics
6.4.1 Marginal Revenue and Marginal Cost
Sometimes in economics, we are interested in the effect on total revenue, TR,
of a change in the value of Q. To do this, the concept of marginal revenue is
introduced. The marginal revenue of a good, MR, is defined by
MR = d(TR)
dQ
.
The marginal revenue function measures the instantaneous rate of change in total
revenue, TR, compared with demand, Q. For example, the marginal revenue
function, MR, corresponding to
TR = 100Q − 2Q2
is given by
MR = d(TR)
dQ
= 100 − 4Q.
If the current demand is 15, say, then
MR = 100 − 4 × 15 = 40.
This means that when demand is changed slightly from its current value of 15,
the corresponding change in total revenue is 40 times as large. However, if the
demand is 20, then
MR = 100 − 4 × 20 = 20,
which means that when demand is changed slightly from Q = 20, the corresponding
change in total revenue is only 20 times as large.
Economists say that MR is approximately the change in TR resulting from
a one unit increase in demand Q. In general,
Δ(TR) ≈ MR× ΔQ.
(This is just a consequence of the small increments formula (6.4).) This approximation
is a good one provided the quantities of Q involved are very large
122 Elements of Mathematics for Economics and Finance
so that one unit is relatively very small. An analogous statement can be made
regarding marginal cost, MC.
The marginal cost function, MC, is defined by
MC = d(TC)
dQ
. (6.13)
The average cost function, AC, is defined by
AC = TC
Q
. (6.14)
Example 6.5
If the average cost function for a good is
AC =
24
Q
- 15 + 3Q,
find an expression for the total cost function. What are the fixed costs in this
case? Write down an expression for the marginal cost function.
Solution. To find an expression for TC, we use the formula for AC given by
(6.14). Hence
TC = AC × Q
24
Q
- 15 + 3Q
Q
= 24+15Q + 3Q2.
Since TC = FC + (V C)Q, the fixed cost element of the total cost function is
independent of Q. Therefore, in this example the fixed costs are 24. Finally,
an expression for the marginal cost function is obtained by differentiating TC
with respect to Q. Therefore,
MC = d(TC)
dQ
= 15+6Q.
Note that the fixed costs have no influence on the marginal cost function since
the derivative of a constant is zero.
- Differentiation 123
6.4.2 Marginal Propensities
The relationship between consumption C and national income Y is sometimes
of the form
C = f(Y ),
where f is some appropriate consumption function. Of interest is the effect on
C due to variations in Y , i.e., if national income rises by a certain amount,
what effect does this have on the spending patterns of the population. This
is analyzed using the concept of marginal propensity to consume, MPC,
defined by
MPC = dC
dY
,
i.e., the marginal propensity to consume is the derivative of consumption with
respect to income. For example, if the consumption function is
C = 0.01Y 2 + 0.2Y + 50
to calculate MPC when Y = 30, we have
MPC = dC
dY
= 0.02Y + 0.2.
When Y = 30, MPC = (0.02)(30) + 0.2 = 0.8.
Economists say that MPC is approximately the change in consumption due
to a one unit increase in national income Y . More generally, if national income
increases by a small amount ΔY , then the corresponding small change ΔC in
consumption is approximately MPC × ΔY , i.e.,
ΔC ≈ MPC × ΔY.
If national income is used up only in consumption and savings, then
Y = C + S.
If we differentiate both sides of this equation with respect to Y :
dY
dY
= dC
dY
- dS
dY
,
i.e.,
1 = MPC + MPS,
where
MPS = dS
dY
is the marginal propensity to save. Economists say that MPS is approximately
the change in savings due to a one unit increase in national income Y .
124 Elements of Mathematics for Economics and Finance
More generally, we can show, using the small increments formula again, that
if national income increases by a small amount ΔY , then the corresponding
small change ΔS in savings is given by
ΔS ≈ MPS × ΔY.
Thus if we knowMPC, we can easily determineMPS. In the above example
the value of MPS when Y = 30 is given by
1 = 0.8 + MPS
i.e.,
MPS = 0.2.
This indicates that when income increases by one unit (from its current level of
30), consumption rises by approximately 0.8 units, whereas savings rise by approximately
0.2 units. At this level of income, the nation has a greater propensity
to consume than it has to save.
Example 6.6
If the consumption function is
C = 0.005Y 2 + 0.3Y + 20,
calculate the marginal propensities to consume and save when Y = 10 and give
an interpretation of the results.
Solution. The marginal propensity to consume is defined by
MPC = dC
dY
= 0.01Y + 0.3.
When Y = 10,
MPC = 0.01 × 10 + 0.3 = 0.1 + 0.3 = 0.4.
If national income is used up in consumption and savings only, then
MPC +MPS = 1.
When Y = 10, the marginal propensity to save is
MPS = 1−MPC = 1− 0.4 = 0.6.
Therefore, at this level of national income, the nation has a greater propensity
to save than it has to consume.
- Differentiation 125
6.5 Approximation to Marginal Functions
The exact value of MR at the point Q0 is given by
d(TR)
dQ
,
and so is given by the slope of the tangent to the total revenue function at A
(see Fig. 6.2). The point B also lies on the curve – it corresponds to a one unit
increase in Q, i.e., ΔQ = 1. The vertical distance from A to B therefore equals
the change in TR when Q increases by one unit. The slope of the chord joining
A to B is
Δ(TR)
ΔQ
= Δ(TR)
1
= Δ(TR).
Note that the slope of the tangent is approximately the same as that of the
chord joining A and B. Therefore, the latter produces a reasonable approximation
to MR in many cases.
This approximation holds for any value of ΔQ. Therefore, as we have seen
in Section 6.4.1
MR ≈ Δ(TR)
ΔQ
, (6.15)
or
Δ(TR) ≈ MR× ΔQ, (6.16)
i.e., change in total revenue ≈ marginal revenue × change in demand. Note
that if the total revenue function is linear, then we have equality: Δ(TR) =
MR× ΔQ.
TR
Q
A
B
tangent
Qo Qo+ΔQ
ΔQ
Δ(TR)
Figure 6.2 Approximation to marginal revenue.
126 Elements of Mathematics for Economics and Finance
Example 6.7
If the total revenue function of a good is given by
100Q − Q2,
write down an expression for the marginal revenue function. If the current
demand is 60, estimate the change in the value of TR due to a two unit increase
in Q.
Solution. To determine the marginal revenue function, we differentiate the
total revenue function. Therefore,
MR = d(TR)
dQ
= 100 − 2Q.
When Q = 60,
MR = 100 − 2 × 60 = 100 − 120 = −20.
When there is a two unit increase in Q, i.e., ΔQ = 2, then the estimated change
in TR is given by
Δ(TR) ≈ MR× ΔQ = −20 ×2 = −40,
i.e., there is an estimated 40 unit reduction in TR.
A similar approximation to (6.15), using the small increments formula (6.4),
holds for the marginal cost function:
MC ≈ Δ(TC)
ΔQ
, (6.17)
or
Δ(TC) ≈ MC × ΔQ, (6.18)
i.e., change in total cost ≈ marginal cost × change in demand. Note that we
have equality if the total cost function is linear, then Δ(TC) = MC × ΔQ.
Example 6.8
Find the marginal cost function given the average cost function
AC =
100
Q
- 2.
Deduce that a one unit increase in Q will always result in a two unit increase
in TC, irrespective of the current level of output.
- Differentiation 127
Solution. To determine the marginal cost function, it is first necessary to find
an expression for the total cost function, TC. Now TC and AC are related by
AC = TC
Q
,
and therefore
TC = AC × Q =
100
Q
- 2
Q = 100 + 2Q.
The corresponding marginal cost function is
MC = d(TC)
dQ
= 2.
Since TC is a linear function, we have Δ(TC) = MC × ΔQ. Therefore, if
output increases by one unit, i.e., ΔQ = 1, then
Δ(TC) = 2,
irrespective of the current level of output.
6.6 Higher Order Derivatives
We have already seen that the derivative of a function of x is itself a function of
x. This suggests the possibility of differentiating a second time to get the ‘slope
of the slope of a function’. This is written as f(x) or d2y/dx2. This function
is known as the second order derivative of f(x). Higher order derivatives are
found by applying the rules of differentiation to lower order derivatives. The
third order derivative f(x) or d3y/dx3 measures the slope and rate of change
of the second order derivative, etc. Thus, if
f(x) = 2×4 + 5×3 + 3×2,
we have
f
(x) = 8×3 + 15×2 + 6x
f
(x) = 24×2 + 30x + 6
f
(x) = 48x + 30
f(4)(x) = 48
f(5)(x) = 0
128 Elements of Mathematics for Economics and Finance
Example 6.9
For each of the following functions, find the second derivative and evaluate it
at x = 2.
- f(x) = x6 + 3×4 + x,
- y = 2×2 + 38x − 6,
- y = (8x − 4)6.
Solution. - To differentiate this polynomial function, we use a combination of the rule
for the sum of functions and the power function rule. So
f
(x) = 6×6−1 + 3 × 4×4−1 + 1
= 6×5 + 12×3 + 1.
Differentiating a second time gives
f
(x) = 6× 5×5−1 + 12 × 3×2−1 + 0
= 30×4 + 36×2
Evaluating the second derivative when x = 2, we have
f
(2) = 30(24) + 36(22) = 624 - To differentiate this quadratic function, we use a combination of the rule
for the sum of functions and the power function rule. So
dy
dx
= 2× 2×2−1 + 38
= 4x + 38.
Differentiating a second time gives
d2y
dx2 = 4
At x = 2, d2y
dx2 = 4. - To differentiate this function, we use the chain rule. Let u = 8x − 4, then
y = u6. Since
dy
du
= 6u5, and du
dx
= 8, - Differentiation 129
we have, using the chain rule
dy
dx
= dy
du
du
dx
= 6u5 × 8 = 48(8x − 4)5.
Applying the chain rule a second time gives
d2y
dx2 = 48 × 5u4 × 8 = 1920(8x − 4)4.
Evaluating the second derivative when x = 2 gives
d2y
dx2 = 39813120.
6.7 Production Functions
In one of the simplest models for production, the quantity of output produced,
Q, is assumed to be a function of capital, K, and labour, L. However, in the
short run K can be assumed to be fixed and so Q is then a function of L alone.
In this instance, Q is referred to as the short run production function. The
independent variable L is usually measured in terms of the number of workers
or the number of worker hours. The derivative of the production function with
respect to L, known as the marginal product of labour (MPL), measures
the rate at which output changes as the number of workers increases. Thus, we
have
MPL = dQ
dL
. (6.19)
Economists say that MPL is approximately the change in output resulting from
a one unit increase in labour.
Example 6.10
For the production function
Q = 8
√
L,
find the marginal product of labour. Determine the output and the marginal
product of labour when - L = 1,
- L = 4,
- L = 100.
130 Elements of Mathematics for Economics and Finance
20 40 60 80 100
0
20
40
60
80 Q
L
Figure 6.3 Graph of the production function Q = 8L1/2.
Solution. The marginal product of labour is found by differentiating Q =
8L1/2. This gives, using the power function rule,
MPL = dQ
dL
= 8× 1
2L1/2−1 = 4L
−1/2 =
4
L1/2 . - When L = 1, Q = 8 and MPL = 4.
- When L = 4, Q = 16 and MPL = 2.
- When L = 100, Q = 80 and MPL = 0.4.
As L increases from 0, so does output (see Fig. 6.3). However, MPL decreases
and therefore although output increases, it does so at a decreasing rate. In this
situation, we say that there are diminishing returns to labour.
Example 6.11
Consider the production function is
Q = 120
√
L − 5L,
where Q denotes output and L denotes the size of the workforce. Calculate the
value of MPL when - L = 1,
- L = 16,
- L = 100,
- L = 900,
and discuss the implication of these results. - Differentiation 131
50 100 150
0
50
100
150
Q
L
Figure 6.4 Graph of the production function Q = 60
√
L − 5L.
Solution. The marginal product of labour, MPL, is found by differentiating
the production function with respect to L. This gives
MPL = dQ
dL
= 120 × 1
2L1/2−1 − 5 = 60L
−1/2 −5 =
60
L1/2
− 5. - When L = 1, MPL = 55.
- When L = 16, MPL = 10.
- When L = 100, MPL = 1,
- When L = 900, MPL = −3.
In the last part of this example, we see that a size of workforce is reached that,
if exceeded, actually results in a decrease in output. This may seem counterintuitive
at first sight. However, this situation can occur in production processes
where productivity is diminished due to problems of overcrowding on the shop
floor or the need to create an elaborate administration to organize the larger
workforce. The graph of this production function is sketched in Fig. 6.4.
The production function in the last example satisfies what is known as the
law of diminishing marginal productivity. This law, also known as the law
of diminishing returns, states that the increase in output due to a one unit
increase in labour will eventually decline. A typical production function that
satisfies this law is shown in Fig. 6.5. The graph of the corresponding marginal
product of labour, MPL, is shown in Fig. 6.6. Note that the maximum value of
MPL is attained when L = L0 and MPL = 0 at the value of L corresponding
to maximum production.
132 Elements of Mathematics for Economics and Finance
A
B
C D
Q
Lo L
Figure 6.5 Graph illustrating a production function that satisfies the law
of diminishing marginal productivity.
Between L = 0 and L = L0, the curve bends upwards, becoming progressively
steeper and so the slope, MPL, of the production function increases.
Mathematically speaking,
d(MPL)
dL
0,
MP
Lo L
L
Figure 6.6 Graph of marginal product of labour corresponding to the
production function shown in Fig. 6.5.
- Differentiation 133
or, since MPL = dQ/dL,
d2Q
dL2 > 0,
i.e., if we take any two points A: (L1,Q(L1)) and B: (L2,Q(L2)) on the curve
between the points (0, 0) and (L0,Q(L0)) with L1 < L2, then the slope of the tangent at B is greater than that at A (see Fig. 6.5). Similarly, for L > L0 the
curve of the production function bends downwards and the slope of the slope
function decreases and is negative, i.e.,
d2Q
dL2 < 0, i.e., if we take any two points C: (L3,Q(L3)) and D: (L4,Q(L4)) on the curve beyond the point (L0,Q(L0)) with L3 < L4, then the slope of the tangent at C is greater than that at D (see Fig. 6.5). The law of diminishing returns states that this must happen eventually, i.e., d2Q dL2 < 0, for L > L0.
Example 6.12
Show that the law of diminishing marginal productivity holds for the production
function
Q = 15L2 − 0.2L3.
Solution. Differentiating the production function gives
MPL = dQ
dL
= 30L − 0.6L2.
Differentiating a second times gives
d2Q
dL2 = 30 − 1.2L.
The expression defining the second derivative, i.e., 30−1.2L becomes negative
when 30 − 1.2L < 0, i.e., when L >
30
1.2
= 25.
Therefore, the law of diminishing marginal productivity holds for this production
function for L > 25, i.e., L0 = 25.
134 Elements of Mathematics for Economics and Finance
EXERCISES
6.1. Find the derivatives of the following functions:
a) f(x) = 4,
b) f(x) = 4×3,
c) f(x) = x8,
d) f(x) = 2×3/2,
e) f(x) = 3x + 7.
6.2. Find dy/dx for each of the following:
a) y = 5+2x − 3×2,
b) y = x3 + 3×2 + 5,
c) y = x2 + 5,
d) y = x4 − 3×2 + 1.
6.3. Find the first and second derivatives of the following functions:
a) y = e4x,
b) y = 3e−2x.
Evaluate these derivatives when x = 0.
6.4. Find the first and second derivatives of the following functions:
a) y = ln4x,
b) y = 2ln7x.
Evaluate these derivatives when x = 1.
6.5. If TC = 3Q2+7Q+12, find expressions for the marginal and average
cost functions. Evaluate them when Q = 3 and Q = 5.
6.6. For each of the following demand functions, find expressions for TR
and MR and evaluate them when Q = 4 and Q = 10.
a) Q = 36 − 2P,
b) 44 − 4P − Q = 0.
6.7. Find the first and second derivatives of the function
y = 6×3 − 20×2 − 9x + 12.
Evaluate these derivatives when x = 1. - Differentiation 135
6.8. Find the marginal cost function for the average cost function given
by
AC = 3
2Q+4+ 46
Q .
6.9. The fixed costs of producing a good are 50 and the variable costs
are 2 + 1
4Q per unit.
a) Find expressions for TC and MC.
b) Evaluate TC and MC when Q = 20. Hence estimate the change
in TC brought about by a 2 unit increase in output from the
current level of 20 units.
6.10. If the consumption function is
C = 0.03Y 2 + 0.1Y + 30
calculate MPC and MPS when Y = 4 and give an interpretation
of the results.
6.11. Show that the law of diminishing marginal productivity holds for
the production function
Q = 18L2 − 0.6L3.
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