Topic 4: Differentiation



Topic 5: Differentiation

Lecture Notes:

section 4

Jacques Text Book (edition 3):

Chapter4

Recall measuring change in the case of a linear function:

y = a + bx

a = intercept

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

b = [pic] = [pic]

( constant along a straight line

( y changes at a constant rate in response to changes in x

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

[pic] = [pic] 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

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

- which is different for different values of x

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

[pic] = 2x+(x

The slope depends on x and (x

Differentiation: finds the derived function by letting change in x become arbitrarily small, i.e. letting ( x ( 0

[pic] = 2x in the limit, as (x (0

[pic]

Rules for Differentiation (section 4.3)

1. The Constant Rule

If y = c where c is a constant,

[pic]

e.g. y = 10 then [pic]

2. The Linear Function Rule

If y = a + bx

[pic]

e.g. y = 10 + 6x then [pic]

3. The Power Function Rule

If y = axn, a & n are constants

[pic]

i) y = 4x => [pic]

ii) y = 4x2 => [pic]

iii) y = 4x3 => [pic]

iv) y = 4x-2 => [pic]

4. The Sum-Difference Rule

If y = f(x) ( g(x)

[pic]

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

[pic]

(ii) y = 4x2 - x3 - 4x then [pic]

(iii) y = 5x + 4 then [pic]

5. The Product Rule

If y = u.v where u and v are functions of x

Then [pic]

i) y = (x+2)(ax2+bx)

[pic]

ii) y = (4x3-3x+2)(2x2+4x)

[pic]

6. The Quotient Rule

If y = u/v where u and v are functions of x

Then [pic]

i) y = (x+2)/(x+4)

[pic]

ii) y = (3x+2)/(x2+4)

[pic]

[pic]

7. The Chain Rule

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

[pic]

i) y = (ax2 + bx)½

let v = (ax2 + bx) , so y = v½

[pic]

ii) y = (4x3 + 3x – 7 )4

let v = (4x3 + 3x – 7 ), so y = v4

[pic]

8. The Inverse Function Rule

If x = f(y) then [pic]

The derivative of the inverse of the function x = f(y), is the inverse of the derivative of the function

(i) x = 3y2 then

[pic] so [pic]

(ii) y = 4x3 then

[pic] so [pic]

1. Differentiating functions using Rules 1 ( 8,

See Section 4 of course manual, questions 3, 4 and 10

Applications of the Basic Rules

Calculating Marginal Functions

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

Example 2:

If a firms Total Cost Curve is:

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

(vi) Suppose now TC=Q3- 4Q2+12Q +10. Draw new curves and comment….

1) Find the Average Cost

AC = TC / Q = Q2 – 4Q + 12

2) Find the Marginal Cost

[pic]

3) When does AC = MC?

Q2 – 4Q + 12 = 3Q2 – 8Q + 12

( 2Q2 – 4Q = 0

( 2Q = 4

( Q = 2

Thus, AC = MC curves when Q = 2

4) When does the slope of AC = 0?

Differentiate AC = Q2 – 4Q + 12 to find slope……

[pic]

then set it equal to 0

2Q – 4 = 0

( Q = 2 when slope AC = 0

(v) Economic Significance?

MC curve cuts the AC curve at its minimum point…….(draw both curves)

MC cuts AC curve at minimum point…

(vi) What happens if we introduce Fixed costs to the TC function?

TC=Q3- 4Q2+12Q +10

( no impact on the MC function,

( shift up AC function by FC/q

Example 3: ELASTICITY

Price Elasticity of Demand:

ed = [pic]

= [pic] = [pic]

To calculate the point elasticity of demand then,

ed = [pic]

e.g. Find ed of the function Q = aP-b

ed = [pic]

= [pic]

Inelastic demand: if (ed( < 1

Unit elastic demand: if (ed( = 1

Elastic demand: if (ed( > 1

9. Differentiating Exponential Functions (Course Manual, parts of Topic 6.1)

Aside: The exponential function:

y = exp(x) = ex

Features of y = ex

1. non-linear

2. always positive

3. as ( x get ( y and ( slope of graph

exponential function can be differentiated

Rule 9:

If y = ex then [pic] where e = 2.71828….

More generally,

If y = Aerx then [pic]

Examples:

1) y = e2x then [pic] = 2e2x using above rule

2) y = e-7x then [pic] = -7e-7x

.Differentiating Natural Logs

(Course Manual, Topic 6.2)

Thus, if y = ex then x = loge y = ln y

Logs to the base e are natural logs

Differentiating Natural Logs

5. If y = ex then [pic] = y

6. From The Inverse Function Rule

y = ex ( [pic]

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

8. Thus, x = ln y ([pic]

Rule 9: Differentiating Natural Logs

if y = loge x = ln x ( [pic]

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

Thus, if y = ln mx ( [pic]

Proof

if y = ln mx m>0

Rules of Logs ( y = ln m+ ln x

Differentiating (Sum-Difference rule)

[pic]

Examples:

1) y = ln 5x (x>0) ( [pic]

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

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

Chain Rule: ( [pic]

[pic]

[pic]

3) y = x4lnx

Product Rule: (

[pic]

= [pic] = [pic]

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

Simplify first using rules of logs

( y = lnx3 + ln(x+2)4

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

[pic]

Note:

2. Differentiating exponential and log functions using Rules 9 and 10,

See Section 6 of course manual, questions 3 and 4**

Example 1

If the Demand equation is given by

P = 200 – 40ln(Q+1)

Calculate the price elasticity of demand when Q = 20

Solution

Price elasticity of demand:

ed = [pic] ................
................

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