The learning outcomes of this topic are: - Carry out partial differentiation - Relate partial differentiation to optimization - Calculate partial point elasticities - Recognize the total differential This topic will cover: - Partial Differentiation - Total differential

1. Topic 6:
Differentiation 2

2. This topic will cover:
◦ Partial Differentiation
◦ Total differential

3. By the end of this topic students will be able
to:
◦ Carry out partial differentiation
◦ Relate partial differentiation to optimization
◦ Calculate partial point elasticities
◦ Recognise the total differential

4. ◦ In each case differentiate y with respect to S.
𝑦 = −4𝑥2 + 50𝑥
𝑑𝑦
𝑑𝑥
= −8𝑥 + 50
𝑦 = 2𝑥 + 5 3
𝑦 = ea𝑥
𝑦 = a 𝑥
𝑦 = 𝑥 𝑥
𝑑𝑦
𝑑𝑥
= 6 2𝑥 + 5 2
𝑑𝑦
𝑑𝑥
= aea𝑥
𝑑𝑦
𝑑𝑥
= a 𝑥 𝑙𝑛a
𝑑𝑦
𝑑𝑥
= 𝑥 𝑥(1 + 𝑙𝑛 𝑥 )

5. ◦ Functions with more than one independent variable
 y = f (x1, x2, x3 …)
 For example:
z = -4x2 – 10y2 + 50x + 100y
0 1 2 3 4 5 6 7 8 9 10
0
100
200
300
400
0
2
4
6
8
10

6. ◦ Partial differentiation
 Differentiate with respect to one variable, assuming all
others are constant
◦ Notation

𝜕𝑧
𝜕𝑥
rate of change of z with x, other variables
assumed constant

𝜕𝑧
𝜕𝑦
rate of change of z with y, other variables
assumed constant

7. •z = -4x2 – 10y2 + 50x + 100y
𝜕𝑧
𝜕𝑥
= −8𝑥 + 50
𝜕𝑧
𝜕𝑦
= −20𝑦 + 100
0 1 2 3 4 5 6 7 8 9 10
0
100
200
300
400
0
2
4
6
8
10

8. •Find partial derivatives with respect to x and y.
𝜕𝑧
𝜕𝑥
= 16𝑥 + 𝑦
𝜕𝑧
𝜕𝑦
= 12𝑦2 + 𝑥𝑧 = 8𝑥2 + 4𝑦3 + 𝑥𝑦
𝑧 = 𝑥2 𝑒3𝑦
𝜕𝑧
𝜕𝑥
= 2𝑥𝑒3𝑦
𝜕𝑧
𝜕𝑦
= 3𝑥2 𝑒3𝑦

9. •A company sells two products, and has
developed a model of its annual profits based
upon the product prices. How should the
prices be set?
0 1 2 3 4 5 6 7 8 9 10
0
100
200
300
400
0
2
4
6
8
10
𝜋 = −4𝑝1
2
− 10𝑝2
2
+ 50𝑝1 + 100𝑝2

10. 0 1 2 3 4 5 6 7 8 9 10
0
100
200
300
400
0
2
4
6
8
10
𝜋 = −4𝑝1
2
− 10𝑝2
2
+ 50𝑝1 + 100𝑝2
𝜕𝜋
𝜕𝑝1
= −8𝑝1 + 50
𝜕𝜋
𝜕𝑝2
= −20𝑝2 + 100
= 0
= 0
−8𝑝1 + 50 = 0 ⟹ 𝑝1 = 6.25
−20𝑝2 + 100 ⟹ 𝑝2 = 5
•Hence maximum profit is 406.25

11. ◦ Elasticity of demand
𝐸 𝑤 =
% change in 𝑄
% change in 𝑤
=
Δ𝑄
𝑄
𝑤
Δ𝑤
◦ Arc price elasticity of demand
𝐸 𝑤 =
(𝑄2−𝑄1)
𝑄2 + 𝑄1
2
(𝑤2+𝑤1)
2
(𝑤2−𝑤1)
◦ Point elasticity of demand
𝐸 𝑤 = lim
Δ𝑄,Δ𝑤⟶0
Δ𝑄
Δ𝑤
𝑤
𝑄
=
𝜕𝑄
𝜕𝑤
𝑤
𝑄
(Q1,w1)
(Q2,w2)
quantity Qvariablew

12. ◦ Own price elasticity of demand
𝐸 𝑝 =
𝜕𝑄1
𝜕𝑝1
𝑝1
𝑄1
◦ Cross price elasticity of demand
𝐸12 =
𝜕𝑄1
𝜕𝑝2
𝑝2
𝑄1
◦ Income elasticity of demand
𝐸𝐼 =
𝜕𝑄1
𝜕𝐼
𝐼
𝑄1

13. ◦ Following research the demand for Brand 1 is thought
to be characterised by;
𝑄1 = 50 − 2𝑝1 + 𝑝2 + 3𝑝3 + 0.01𝐼 + other terms
◦ What is the responsiveness of the quantity demanded
to changes in own price, other brands’ prices and
income?
𝐸p =
𝜕𝑄1
𝜕𝑝1
𝑝1
𝑄1
, 𝐸12 =
𝜕𝑄1
𝜕𝑝2
𝑝2
𝑄1
, 𝐸13 =
𝜕𝑄1
𝜕𝑝3
𝑝3
𝑄1
, 𝐸𝐼 =
𝜕𝑄1
𝜕𝐼
𝐼
𝑄1

14. 𝑄1 = 50 − 2𝑝1 + 𝑝2 + 3𝑝3 + 0.01𝐼 + other terms
𝐸 𝑝 =
𝜕𝑄1
𝜕𝑝1
𝑝1
𝑄1
𝐸12 =
𝜕𝑄1
𝜕𝑝2
𝑝2
𝑄1
𝐸13 =
𝜕𝑄1
𝜕𝑝3
𝑝3
𝑄1
𝐸 𝑝 = −2
𝑝1
𝑄1
𝐸12 =
𝑝2
𝑄1
𝐸13 = 3
𝑝3
𝑄1
𝐸𝐼 = 0.01
𝐼
𝑄1
𝐸𝐼 =
𝜕𝑄1
𝜕𝐼
𝐼
𝑄1

15. 𝑄1 = 100 − 4𝑝1 + 2𝑝2
𝐸 𝑝 =
𝜕𝑄1
𝜕𝑝1
𝑝1
𝑄1
= −4
𝑝1
𝑄1
= −1
𝐸12 =
𝜕𝑄1
𝜕𝑝2
𝑝2
𝑄1
= 2
𝑝2
𝑄1
= 0.33
•What are the own price and cross price point
elasticities at p1 = £15, p2 = £10?

16. ◦ Total Differential
𝑦 = 𝑦(𝑥1 + 𝑥2 + 𝑥3 +...)
𝑑𝑦 =
𝜕𝑦
𝜕𝑥1
𝑑𝑥1 +
𝜕𝑦
𝜕𝑥2
𝑑𝑥2 +
𝜕𝑦
𝜕𝑥3
𝑑𝑥3 +...
∆𝑦 ≈
𝜕𝑦
𝜕𝑥1
∆𝑥1 +
𝜕𝑦
𝜕𝑥2
∆𝑥2 +
𝜕𝑦
𝜕𝑥3
∆𝑥3 +...

17. ◦ Show that for the demand function Q1=Q1(p1,p2) that
% increase
in
demand
for Brand
1
own price
elasticity
of
demand
% increase
in price of
Brand 1
cross
price
elasticity
of
demand
% increase
in price of
Brand 2
= +× ×

18. % increase
in
demand
for Brand
1
own price
elasticity
of
demand
% increase
in price of
Brand 1
cross
price
elasticity
of
demand
% increase
in price of
Brand 2
= +× ×
∆𝑄1 ≈
𝜕𝑄1
𝜕𝑝1
∆𝑝1 +
𝜕𝑄1
𝜕𝑝2
∆𝑝2
∆𝑄1 ≈ 𝐸 𝑝 𝑄1
∆𝑝1
𝑝1
+ 𝐸12 𝑄1
∆𝑝2
𝑝2
∆𝑄1
𝑄1
≈ 𝐸 𝑝 ×
∆𝑝1
𝑝1
+ 𝐸12 ×
∆𝑝2
𝑝2

19. By the end of this topic students will be able
to:
◦ Carry out partial differentiation
◦ Relate partial differentiation to optimization
◦ Calculate partial point elasticities
◦ Recognise the total differential

20. Any Questions?