Calculus for Economics: Derivatives and Their Applications | Study notes Economics

CHAPTER 5

Differentiation

Differentiation is a technique that enables us to find out how a function

changes when its argument changes. It is an essential tool in economics.

If you have done A-level maths, much of the maths in this chapter will

be revision, but the economic applications may be new. We use the first

and second derivatives to work out the shapes of simple functions,

and find maximum and minimum points. These techniques are

applied to cost, production and consumption functions.Concave

and convex functions are important in economics.

—./—









1. What is a Derivative?

1.1. Why We Are Interested in the Gradient of a Function

In many economic applications we want to know how a function changes when its argument

changes – so we need to know its gradient.

This graph shows the production function Y(L)

of a firm: if it employs Lworkers it produces Y

units of output.

gradient = ∆Y

∆L=200

10 = 20

The gradient represents the marginal product of

labour - the firm produces 20 more units of out-

put for each extra worker it employs.

The steeper the production function, the greater

is the marginal product of labour.

Y(L)

∆L

∆Y

10 20 30 40 50

200

400

600

800

With this production function, the gradient

changes as we move along the graph.

The gradient of a curve at a particular point is

the gradient of the tangent.

The gradient, and hence the marginal product

of labour, is higher when the firm employs 10

workers than when it employs 40 workers. .

•

Y(L)

10 20 30 40 50

200

400

600

800

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CHAPTER 5

Differentiation

Differentiation is a technique that enables us to find out how a function

changes when its argument changes. It is an essential tool in economics.

If you have done A-level maths, much of the maths in this chapter will

be revision, but the economic applications may be new. We use the first

and second derivatives to work out the shapes of simple functions,

and find maximum and minimum points. These techniques are

applied to cost, production and consumption functions. Concave

and convex functions are important in economics.

What is a Derivative?

1.1. Why We Are Interested in the Gradient of a Function

In many economic applications we want to know how a function changes when its argument

changes – so we need to know its gradient.

This graph shows the production function Y (L)

of a firm: if it employs L workers it produces Y

units of output.

gradient =

∆Y

∆L

The gradient represents the marginal product of

labour - the firm produces 20 more units of out-

put for each extra worker it employs.

The steeper the production function, the greater

is the marginal product of labour.

.. L

Y

Y (L)

∆L

∆Y

With this production function, the gradient

changes as we move along the graph.

The gradient of a curve at a particular point is

the gradient of the tangent.

The gradient, and hence the marginal product

of labour, is higher when the firm employs 10

workers than when it employs 40 workers.

L

Y

Y (L)

1.2. Finding the Gradient of a Function

If we have a linear function: y(x) = mx + c

we know immediately that its graph is a straight line, with gradient equal to m (Chapter 2).

But the graph of the function: y(x) =

x

is a curve, so its gradient changes. To find the gradient at a particular point, we could try

to draw the graph accurately, draw a tangent, and measure its gradient. But we are unlikely

to get an accurate answer. Alternatively:

Examples 1.1: For the function y(x) =

x

(i) What is the gradient at the point where x = 2?

P

Q

x

y

y(x) =

x

∆x

∆y

An approximation to the gradient at P is:

gradient of P Q =

y(3) − y(2)

For a better approximation take a

point nearer to P :

y(2.5) − y(2)

or much nearer:

y(2.001) − y(2)

So we can see that:

In general the gradient at P is given by: lim

∆x→ 0

∆y

∆x

In this case the gradient at P is 1.

(ii) What is the gradient at the point where at the point where x = 3?

Applying the same method:

y(3.1) − y(3)

y(3.01) − y(3)

y(3.001) − y(3)

lim

∆x→ 0

∆y

∆x

Finding the Derivative of the

Function y = x

n

Although we could use the method in the previous section to find the derivative of any

function at any particular point, it requires tedious calculations. Instead there is a simple

formula, which we can use to find the derivative of any function of the form y = x

n

, at any

point.

If y = x

n

, then

dy

dx

= nx

n− 1

So the derivative of the function x

n

is itself a function: nx

n− 1

and we just have to evaluate this to find the gradient at any point.

Examples 2.1: For the function y = x

(i) Find the derivative.

Applying the formula, with n = 3:

dy

dx

= 3x

(ii) Find the gradient at the points: x = 1, x = −2 and x = 10

When x = 1

dy

dx

= 3 × 1

= 3. So the gradient is 3.

When x = − 2

dy

dx

= 3 × (−2)

When x = 10

dy

dx

= 3 × (10)

Some Notation: We could write this as:

dy

dx

x=

The process of finding the derivative function is called

differentiation

Examples 2.2: Differentiate the following functions.

(i) y(x) = x

dy

dx

= 9x

(ii) Q(P ) = Q

dQ

dP

Q

2 Q

(iii) Y (t) = t

dY

dt

= 1. 2 t

(iv) F (L) = L

dF

dL

L

3 L

(v) y =

x

In this case n = −2:

y = x

dy

dx

= − 2 x

x

We have not proved that the formula is correct, but for two special cases we can verify it:

Examples 2.3: The special cases n = 0 and n = 1

(i) Consider the linear function: y = x

We know (from Chapter 2) that it is a straight line with gradient 1.

We could write this function as: y(x) = x

and apply the formula:

dy

dx

= 1 × x

So the formula also tells us that the gradient of y = x is 1 for all values of x.

(ii) Consider the constant function: y(x) = 1

We know (Chapter 2) that it is a straight line with gradient zero and y-intercept 1.

We could write this function as: y(x) = x

and apply the formula:

dy

dx

= 0 × x

So the formula also tells us that the gradient is zero.

Exercises 5.2: Using the formula to find derivatives

(1)(1) If y = x

, find

dy

dx

and hence find the gradient of the function when x = − 2

(2) Find the derivative of y = x

. What is the gradient at the point where x = 16?

(3) If y = x

what is

dy

dx

x=

(4) Find the derivative of the function z(t) = t

and evaluate it when t = 3

(5) Differentiate the supply function Q(P ) = P

(6) Differentiate the utility function u(y) = y

and hence find its gradient when y = 1

(7) Find the derivative of the function y =

x

, and the gradient when x = 2.

(8) Differentiate the demand function Q =

P

What happens to the derivative (a) as P → 0 (b) as P → ∞?

The simple rule for differentiating y = x

n

can be easily extended so that we can differentiate

other functions, such as polynomials. First it is useful to have some alternative notation:

The derivative of the function y(x) can be written as

y

(x) instead of

dy

dx

Examples 4.1: For the function y(x) = x

, we can write:

Either:

dy

dx

= 2x and

dy

dx

x=

Or: y

(x) = 2x and y

You can see that the notation y

(x) (read “y prime x”) emphasizes that the derivative is a

function, and will often be neater, especially when we evaluate the derivative at particular

points. It enables us to write the following rules neatly.

Some rules for differentiation

If f (x) = a f

(x) = 0

If f (x) = ag(x) f

(x) = ag

(x)

If f (x) = g(x) ± h(x) f

(x) = g

(x) ± h

(x)

(a is a constant, and f , g and h are functions.)

The first rule is obvious: the graph of a constant function is a horizontal line with zero

gradient, so its derivative must be zero for all values of x. You could prove all of these rules

using the method of differentiation from first principles in the previous section.

Examples 4.2: Differentiate the following functions.

(i) f (x) = 7

By the first rule:

f

(x) = 0

(ii) y(x) = 4x

We know the derivative of x

is 2x, so using the second rule:

y

(x) = 4 × 2 x = 8x

(iii) z(x) = x

− x

Here we can use the third rule:

z

(x) = 3x

− 4 x

(iv) g(x) = 6x

+ 3x + 5

g

(x) = 6 × 2 x + 3 × 1 + 0 = 12x + 3

(v) h(x) =

x

− 5 x + 4x

h(x) = 7 x

− 5 x + 4x

h

(x) = 7 × (− 3 x

) − 5 + 4 × 3 x

x

− 5 + 12x

(vi) P (Q) = 5Q

− 4 Q + 10

P

(Q) = 6Q

(vii) Y (t) = k

t (where k is a parameter)

Y (t) = kt

Y

(t) = k ×

t

k

2 t

(viii) F (x) = (2x + 1)(1 − x

First multiply out the brackets:

F (x) = 2 x + 1 − 2 x

− x

F

(x) = 2 − 6 x

− 2 x

Examples 4.3:

(i) Differentiate the quadratic function f (x) = 3x

− 6 x + 1, and hence find its gradient at

the points x = 0, x = 1 and x = 2.

f

(x) = 6x − 6, and the gradients are f

(0) = −6, f

(1) = 0, and f

(ii) Find the gradient of the function g(y) = y −

y

when y = 1.

g

(y) = 1 +

y

, so g

Exercises 5.4: Differentiating more complicated functions

(1)(1) If f (x) = 8x

− 7, what is (a) f

(x) (b) f

(2) (c) f

(2) Differentiate the functions (a) u(y) = 10y

− y

(b) v(y) = 7y + 5 − 6 y

(3) If Q(P ) =

P

, what is Q

(P )?

(4) Find the gradient of h(z) = z

(z + 4) at the point where z = 2.

(5) If y = t − 8 t

, what is

dy

dt

(6) Find the derivative of g(x) = 3x

+ 2 − 8 x

and evaluate it when x = 1.

(7) Differentiate y(x) = 12x

+ 7x

− 4 x

− 2 x + 8

(8) If F (Y ) = (Y − 1)

+ 2Y (1 + Y ), what is F

The derivative of the total cost function C(Q) is the

marginal cost: MC =

dC

dQ

Examples 5.2: Consider the following total cost functions:

(i) C(Q) = 4Q + 7

Differentiating the cost function: MC=4. So marginal cost is constant for all levels of

output Q. The cost function is an upward-sloping straight line.

(ii) C(Q) = 3Q

+ 4Q + 1

In this case, MC=6Q + 4. Looking at this expression for MC, we can see that the firm

has increasing marginal cost – the higher the level of output, the more it costs to make

an additional unit.

Q Q

C C

(i) Constant marginal cost (ii) Increasing marginal cost

5.3. Consumption Functions

In macroeconomics the consumption function C(Y ) specifies how aggregate consumption C

depends on national income Y.

The derivative of the consumption function C(Y ) is the

marginal propensity to consume:

MPC =

dC

dY

The marginal propensity to consume tell us how responsive consumer spending would be if,

for example, the government were increase income by reducing taxes.

Exercises 5.5: Economic Applications

(1)(1) Find the marginal product of labour for a firm with production function

Y (L) = 300L

. What is the MPL when it employs 8 units of labour?

(2) A firm has cost function C(Q) = a + 2Q

k

, where a and k are positive parameters.

Find the marginal cost function. For what values of k is the firm’s marginal cost

(a) increasing (b) constant (c) decreasing?

(3) If the aggregate consumption function is C(Y ) = 10 + Y

, what is (a) aggregate

consumption, and (b) the marginal propensity to consume, when national income

is 50?

6.1. The Sign of the Gradient

Since the derivative f

(x) of a function is the gradient, we can look at the derivative to find

out which way the function slopes - that is, whether the function is increasing or decreasing

for different values of x.

f

(x) > 0 f is increasing at x

f

(x) < 0 f is decreasing at x

Increasing Function Decreasing Function

f

(x) is positive for

all values of x

f

(x) is negative for

all values of x

f

(x) f

(x)

positive negative

f

(x) f

(x)

negative positive

(i) (ii) (iii) (iv)

6.2. Stationary Points

In diagram (iii) there is one point where the gradient is zero. This is where the function

reaches a maximum. Similarly in diagram (iv) the gradient is zero where the function reaches

a minimum.

A point where f

(x) = 0 is called a stationary point, or a

turning point, or a critical point, of the function.

If we want to know where a function reaches a maximum or minimum, we can try to do it

by looking for stationary points. But we have to be careful: not all stationary points are

maxima or minima, and some maxima and minima are not stationary points. For example:

This function has four stationary points:

a local maximum at A and C
a local minimum at B
and a stationary point which is neither

a max or a min at D. This is called a

point of inflection.

The global maximum of the function (over the

range of x-values we are looking at here) is at C.

The global minimum of the function is at E,

which is not a stationary point.

x

f (x)

A

B

C

D

E

Note that there may not be a global maximum and/or minimum; we know that some functions

go towards ±∞. We will see some examples below.

(iii) h(x) =

x

for 0 < x ≤ 1

In this example we are only considering a limited range of values for x.

The derivative is: h

(x) = −

x

There are no stationary points because:

x

< 0 for all values of x between 0 and 1

The gradient is always negative: h is a

decreasing function of x.

So the minimum value of the function is

at x = 1: h(1) = 2.

The function doesn’t have a maximum

value because:

h(x) → ∞ as x → 0.

Exercises 5.6: Stationary Points

For each of the following functions (i) find and identify all of the stationary points, and

(ii) find the global maximum and minimum values of the function, if they exist.

(1)(1) a(x) = x

+ 4x + 2

(2) b(x) = x

− 3 x + 1

(3) c(x) = 1 + 4x − x

for 0 ≤ x ≤ 3

(4) d(x) =

x

+ αx + 1 where α is a parameter.

Hint: consider the two cases α ≥ 0 and α < 0.

Both of these books use the second derivative to identify stationary points, so you may want

to read section 7.2 before looking at them.

The Second Derivative

The derivative y

(x) tells us the gradient, or in other words how y changes as x increases.

But y

(x) is a function.

So we can differentiate y

(x), to find out how y

(x) changes as x increases.

The derivative of the derivative is called the second derivative. It can be denoted y

(x), but

there are alternatives:

For the function y(x) the derivatives can be written:

st

derivative: y

(x) or

d

dx

(y(x)) or

dy

dx

nd

derivative: y

(x) or

d

dx

dy

dx

or

d

y

dx

Examples 7.1: Second Derivatives

(i) f (x) = ax

+ bx + c

The 1

st

derivative is: f

(x) = 2 ax + b

Differentiating again: f

(x) = 2 a

(ii) y =

x

This function can be written as y = 5x

so:

First derivative:

dy

dx

= − 10 x

x

Second derivative:

d

y

dx

= 30 x

x

Exercises 5.7: Find the first and second derivatives of:

(1)(1) f (x) = x

(2) g(x) = x + 1 (3) h(x) = 4x

(4) k(x) = x

(5x

7.1. Using the Second Derivative to Find the Shape of a Function

The 1

st

derivative tells you whether the gradient is positive or negative.

The 2

nd

derivative tells you whether the gradient is increasing or decreasing.

f

(x) > 0 Positive Gradient f

(x) < 0 Negative Gradient

f

(x) > 0 f

(x) < 0 f

(x) > 0 f

(x) < 0

Gradient increasing Gradient decreasing Gradient increasing Gradient decreasing

x = 0: g

(0) = 0 – we can’t tell its type. (We found before

that it is a point of inflection.)

x = 2: g

(2) = − 12 – it is a maximum point.

Exercises 5.8: Find and classify all the stationary points of the following functions:

(1)(1) a(x) = x

− 3 x

(2) b(x) = x +

x

(3) c(x) = x

− 4 x

− 8 x

7.3. Concave and Convex Functions

If the gradient of a function is increasing for all values of x, the function is called convex.

If the gradient of a function is decreasing for all values of x, the function is called concave.

Since the second derivative tells us whether the gradient is increasing or decreasing:

A function f is called

concave

convex

if

f

(x)

0 for all values of x

Convex

Concave

Convex

Examples 7.4: Are the following functions concave, convex, or neither?

(i) The quadratic function f (x) = ax

+ bx + c.

f

(x) = 2a, so if a > 0 it is convex and if a < 0 it is concave.

(ii) P (q) =

q

P

(q) = − 2 q

, and P

(q) = 6q

q

The second derivative is positive for all values of q, so the function is convex.

(iii) y = x

We looked at this function in section 7.1. The second derivative changes sign, so the

function is neither concave nor convex.

7.4. Economic Application: Production Functions

The economic characteristics of a production function Y (L) depend on its shape:

− It is (usually) an increasing function: Y

(L) > 0 for all values of x;

− but the second derivative could be positive or negative: it tells you whether the

marginal product of labour is increasing or decreasing.

L

Y (i)

− Y

(L) < 0 for all values of L

− The production function is concave

− MPL is decreasing

− Diminishing returns to labour

L

Y (ii)

This function is neither concave nor convex.

When employment is low:

− Y

(L) > 0

− Increasing returns to labour

But when emplyment is higher:

− Y

(L) < 0

− Decreasing returns to labour

Exercises 5.9: Concave and Convex Functions in Economics

(1)(1) Consider the cost function C(q) = 3q

+ 2q + 5

(a) What is the marginal cost?

(b) Use the second derivative to find out whether the firm has increasing or

decreasing marginal cost.

(c) Is the cost function convex or concave?

(2) For the aggregate consumption function is C(Y ) = 10 + Y

(a) Find the first and second derivatives.

(b) How does the marginal propensity to consume change as income increases?

(c) Is the function concave or convex?

(d) Sketch the function.

and §4.6 on maxima and minima

Anthony & Biggs §8.2 and §8.3 for stationary points, maxima and minima

Worksheet 5: Differentiation

Quick Questions

(1) Differentiate:

(a) y = 9x

− 7 x

(b) f (x) =

4 x

(c) Y (t) = 100t

(d) P (Q) = Q

− 4 Q

(2) Determine whether each of the following functions is concave, convex, or neither:

(a) y = 5x

− 8 x + 7

(b) C(y) =

4 y (for y ≥ 0)

(c) P (q) = q

− 4 q

(for q ≥ 0)

(d) k(x) = x

− x

(3) Find the first and second derivatives of the production function F (L) = 100L+200L

(for L ≥ 0) and hence determine whether the firm has decreasing, constant, or in-

creasing returns to labour.

(4) Find and classify all the the stationary points of the following functions. Find the

global maxima and minima, if they exist, and sketch the functions.

(a) y = 3x − x

+ 4 for values of x between 0 and 4.

(b) g(x) = 6x

− x for x ≥ 0.

(c) f (x) = x

− 8 x

+ 18x

− 5 (for all values of x)

(5) If a firm has cost function C(Q) = aQ(b + Q

) + c, where a, b and c are positive

parameters, find and sketch the marginal cost function. Does the firm have concave

or convex costs?

Longer Questions

(1) The number of meals, y, produced in a hotel kitchen depends on the number of

cooks, n, according to the production function y(n) = 60n −

n

(a) What is the marginal product of labour?

(b) Does the kitchen have increasing, constant, or decreasing returns to labour?

(c) Find:

(i) the output of the kitchen

(ii) the average output per cook

(iii) the marginal product of labour

when the number of cooks is 1, 5, 10 and 15.

(d) Is this a realistic model of a kitchen? Suggest a possible explanation for the

figures you have obtained.

(e) Draw a careful sketch of the production function of the kitchen.

(f) What is the maximum number of meals that can be produced?

(2) A manufacturer can produce economics textbooks at a cost of £5 each. The text-

book currently sells for £10, and at this price 100 books are sold each day. The

manufacturer figures out that each pound decrease in price will sell ten additional

copies each day.

(a) What is the demand function q(p) for the textbooks?

(b) What is the cost function?

(c) What is the inverse demand function?

(d) Write down the manufacturer’s daily profits as function of the quantity of text-

books sold, Π(q).

(e) If the manufacturer now maximises profits, find the new price, and the profits

per day.

(f) Sketch the profit function.

(g) If the government introduces a law that at least 80 economics textbooks must

be sold per day, and the (law-abiding) manufacturer maximises profits, find the

new price.

Calculus for Economics: Derivatives and Their Applications, Study notes of Economics

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Partial preview of the text

Download Calculus for Economics: Derivatives and Their Applications and more Study notes Economics in PDF only on Docsity!

CHAPTER 5

Differentiation is a technique that enables us to find out how a function

changes when its argument changes. It is an essential tool in economics.

If you have done A-level maths, much of the maths in this chapter will

be revision, but the economic applications may be new. We use the first

and second derivatives to work out the shapes of simple functions,

and find maximum and minimum points. These techniques are

applied to cost, production and consumption functions. Concave

and convex functions are important in economics.

1.1. Why We Are Interested in the Gradient of a Function

In many economic applications we want to know how a function changes when its argument

changes – so we need to know its gradient.

This graph shows the production function Y (L)

of a firm: if it employs L workers it produces Y

units of output.

gradient =

∆Y

∆L

The gradient represents the marginal product of

labour - the firm produces 20 more units of out-

put for each extra worker it employs.

The steeper the production function, the greater

is the marginal product of labour.

.. L

Y

Y (L)

∆L

∆Y

With this production function, the gradient

changes as we move along the graph.

The gradient of a curve at a particular point is

the gradient of the tangent.

The gradient, and hence the marginal product

of labour, is higher when the firm employs 10

workers than when it employs 40 workers.

L

Y

Y (L)

1.2. Finding the Gradient of a Function

If we have a linear function: y(x) = mx + c

we know immediately that its graph is a straight line, with gradient equal to m (Chapter 2).

But the graph of the function: y(x) =

x

is a curve, so its gradient changes. To find the gradient at a particular point, we could try

to draw the graph accurately, draw a tangent, and measure its gradient. But we are unlikely

to get an accurate answer. Alternatively:

Examples 1.1: For the function y(x) =

x

(i) What is the gradient at the point where x = 2?

P

Q

x

y

y(x) =

x

∆x

∆y

An approximation to the gradient at P is:

gradient of P Q =

y(3) − y(2)

For a better approximation take a

point nearer to P :

y(2.5) − y(2)

or much nearer:

y(2.001) − y(2)

So we can see that:

∆x→ 0

∆y

∆x

(ii) What is the gradient at the point where at the point where x = 3?

Applying the same method:

y(3.1) − y(3)

y(3.01) − y(3)

y(3.001) − y(3)

lim

∆x→ 0