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Topic 6: Differentiation

Jacques Text Book (edition 4 ):

Chapter 4

1.Rules of Differentiation

2.Applications

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Differentiation is all about measuring change! Measuring change in a linear function:

y = a + bxa = intercept

b = constant slope i.e. the impact of a unit change in x on the level of y

b = = x

y

12

12xx

yy

3

If the function is non-linear: e.g. if y = x2

0

10

20

30

40

0 1 2 3 4 5 6X

y=

x2

x

y

= 12

12xx

yy

gives slope of the line

connecting 2 points (x1, y1) and (x2,y2) on a curve

(2,4) to (4,16): slope = (16-4)/(4-2) = 6

(2,4) to (6,36): slope = (36-4)/(6-2) = 8

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The slope of a curve is equal to the slope of the line (or tangent) that touches the curve

at that point Total Cost Curve

0

5

10

15

20

25

30

35

40

1 2 3 4 5 6 7

X

y=

x2

which is different for different values of x

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Example:A firms cost function is Y = X2

X X Y Y 0

1

2

3

4

+1

+1

+1

+1

0

1

4

9

16

+1

+3

+5

+7

Y = X2

Y+Y = (X+X) 2

Y+Y =X2+2X.X+X2

Y = X2+2X.X+X2 – Y

since Y = X2 Y = 2X.X+X2

X

Y

= 2X+X

The slope depends on X and X

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The slope of the graph of a function is called the derivative of the

function

• The process of differentiation involves letting the change in x become arbitrarily small, i.e. letting x 0

• e.g if = 2X+X and X 0 = 2X in the limit as X 0

x

y

dx

dyxf

x

0

lim)('

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the slope of the non-linear function

Y = X2 is 2X• the slope tells us the change in y that

results from a very small change in X

• We see the slope varies with X

e.g. the curve at X = 2 has a slope = 4

and the curve at X = 4 has a slope = 8

• In this example, the slope is steeper at higher values of X

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Rules for Differentiation (section 4.3)

1. The Constant Rule

If y = c where c is a constant,

0dx

dy

e.g. y = 10 then 0dx

dy

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2. The Linear Function Rule

If y = a + bx

bdx

dy

e.g. y = 10 + 6x then 6dx

dy

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3. The Power Function Rule

If y = axn, where a and n are constants

1 nx.a.ndx

dy

i) y = 4x => 44 0 xdx

dy

ii) y = 4x2 => xdx

dy8

iii) y = 4x-2 => 38 x

dx

dy

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4. The Sum-Difference Rule If y = f(x) g(x)

dx

)]x(g[d

dx

)]x(f[d

dx

dy

If y is the sum/difference of two or more functions of x:

differentiate the 2 (or more) terms separately, then add/subtract

(i) y = 2x2 + 3x then 34 xdx

dy

(ii) y = 5x + 4 then 5dx

dy

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5. The Product Rule

If y = u.v where u and v are functions of x, (u = f(x) and v = g(x) ) Then

dx

duv

dx

dvu

dx

dy

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Examples

If y = u.v dx

duv

dx

dvu

dx

dy

i ) y = ( x + 2 ) ( a x 2 + b x )

bxaxbaxxdxdy 222

i i ) y = ( 4 x 3 - 3 x + 2 ) ( 2 x 2 + 4 x )

3124244234 223 xxxxxx

dxdy

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6. The Quotient Rule

• If y = u/v where u and v are functions of x (u = f(x) and v = g(x) ) Then

2vdxdv

udxdu

v

dx

dy

15

22 4

2

4

1214

4

2

xx

xx

dx

dy

x

xy

2vdxdv

udxdu

v

dx

dythen

v

uyIf

Example 1

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7. The Chain Rule (Implicit Function Rule)

• If y is a function of v, and v is a function of x, then y is a function of x and

dx

dv.

dv

dy

dx

dy

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Examplesi) y = (ax2 + bx)½

let v = (ax2 + bx) , so y = v½

bax.bxaxdx

dy

2

2

121

2

ii) y = (4x3 + 3x – 7 )4

let v = (4x3 + 3x – 7 ), so y = v4

3127344 233 x.xxdx

dy

dx

dv.

dv

dy

dx

dy

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8. The Inverse Function Rule

• Examples

If x = f(y) then dy

dxdx

dy 1

i) x = 3y2 then

ydy

dx6 so ydx

dy

6

1

ii) y = 4x3 then

212x

dx

dy so 212

1

xdy

dx

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Differentiation in Economics Application I

• Total Costs = TC = FC + VC

• Total Revenue = TR = P * Q = Profit = TR – TC

• Break even: = 0, or TR = TC

• Profit Maximisation: MR = MC

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Application I: Marginal Functions (Revenue, Costs and Profit)

•Calculating Marginal Functions

dQ

TRdMR

dQ

TCdMC

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Example 1

• A firm faces the demand curve P=17-3Q

• (i) Find an expression for TR in terms of Q

• (ii) Find an expression for MR in terms of Q

Solution:

TR = P.Q = 17Q – 3Q2

Q

dQ

TRdMR 617

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Example 2 A firms total cost curve is given by

TC=Q3- 4Q2+12Q

(i) Find an expression for AC in terms of Q

(ii) Find an expression for MC in terms of Q

(iii) When does AC=MC?

(iv) When does the slope of AC=0?

(v) Plot MC and AC curves and comment on the economic significance of their relationship

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Solution

(i) TC = Q3 – 4Q2 + 12Q

Then, AC = TC / Q = Q2 – 4Q + 12

(ii) MC =

1283 2 QQdQ

TCd

(iii) When does AC = MC? Q2 – 4Q + 12 = 3Q2 – 8Q + 12

Q = 2 Thus, AC = MC when Q = 2

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Solution continued….

(iv) When does the slope of AC = 0?

42 QdQ

ACd= 0

Q = 2 when slope AC = 0 (v) Economic Significance? MC cuts AC curve at minimum point…

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9. Differentiating Exponential Functions

If y = exp(x) = ex where e = 2.71828….

then xe

dx

dy

More generally,

If y = Aerx

then ryrAedx

dy rx

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Examples

1) y = e2x then dx

dy = 2e2x

2) y = e-7x then dx

dy = -7e-7x

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10. Differentiating Natural Logs

Recall if y = ex then x = loge y = ln y

If y = ex then xedx

dy = y

From The Inverse Function Rule

y = ex ydy

dx 1

Now, if y = ex this is equivalent to writing x = ln y

Thus, x = ln y ydy

dx 1

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More generally,

if y = ln x xdx

dy 1

NOTE: the derivative of a natural log function does not depend on the co-efficient of x

Thus, if y = ln mx xdx

dy 1

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Proof

if y = ln mx m>0

Rules of Logs y = ln m+ ln x

Differentiating (Sum-Difference rule)

xxdx

dy 110

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Examples1) y = ln 5x (x>0)

xdx

dy 1

2) y = ln(x2+2x+1)

let v = (x2+2x+1) so y = ln v

Chain Rule: dx

dv.

dv

dy

dx

dy

2212

12

x.

xxdx

dy

12

222

xx

x

dx

dy

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3) y = x4lnx

Product Rule:

34 41

x.xlnx

xdx

dy

= xlnxx 33 4 = xlnx 413

4) y = ln(x3(x+2)4)

Simplify first using rules of logs

y = lnx3 + ln(x+2)4

y = 3lnx + 4ln(x+2)

2

43

xxdx

dy

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Applications II

• how does demand change with a change in price……

• ed=

= P

P

Q

Q = Q

P.

PQ

priceinchangealproportion

demandinchangealproportion

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Point elasticity of demand

ed = QP.

dPdQ

ed is negative for a downward sloping demand curve –Inelastic demand if | ed |<1 –Unit elastic demand if | ed |=1 –Elastic demand if | ed |>1

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Example 1

Find ed of the function Q= aP-b

ed = QP.

dPdQ

ed = bb

aP

P.baP

1

= baP

P.

PbaP

b

b

ed at all price levels is –b

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Example 2 If the (inverse) Demand equation is

P = 200 – 40ln(Q+1)

Calculate the price elasticity of demand when Q = 20

Price elasticity of demand: ed = QP.

dPdQ <0

P is expressed in terms of Q,

1

40

QdQ

dP

Inverse rule 40

1

Q

dP

dQ

Hence, ed = QP.

401Q < 0

Q is 20 ed = 2078.22.

4021 = -2.05

(where P = 200 – 40ln(20+1) = 78.22)

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Application III: Differentiation of Natural Logs to find Proportional Changes

The derivative of log(f(x)) f’(x)/f(x), or the

proportional change in the variable x

i.e. y = f(x), then the proportional x

= y.

dx

dy 1 = dx

)y(lnd

Take logs and differentiate to find proportional changes in variables

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1) Show that if y = x, then xy.

dx

dy

1

and this derivative of ln(y) with respect to x.

Solution:

111 x.yy

.dx

dy

= x

x.

y

1

= x

y..

y1

= x

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Solution Continued…

Now ln y = ln x

Re-writing ln y = lnx

xx.

dx

)y(lnd

1

Differentiating the ln y with respect to x gives the proportional change in x.

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Example 2: If Price level at time t is P(t) = a+bt+ct2

Calculate the rate of inflation. Solution:

The inflation rate at t is the proportional change in p

2

21

ctbta

ctb

dt

)t(dP.

)t(P

Alternatively,

differentiating the log of P(t) wrt t directly

lnP(t) = ln(a+bt+ct2)

where v = (a+bt+ct2) so lnP = ln v

Using chain rule,

2

2

ctbta

ctb

dt

)t(Plnd