2. 2
Differentiation is all about measuring change!
Measuring change in a linear function:
y = a + bx
a = intercept
b = constant slope i.e. the impact of a unit
change in x on the level of y
b = =
x
y
1
2
1
2
x
x
y
y
3. 3
If the function is non-linear:
e.g. if y = x2
0
10
20
30
40
0 1 2 3 4 5 6
X
y=x2
x
y
=
1
2
1
2
x
x
y
y
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
4. 4
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
5. 5
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
6. 6
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
dy
x
f
x
0
lim
)
(
'
7. 7
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
9. 9
2. The Linear Function Rule
If y = a + bx
b
dx
dy
e.g. y = 10 + 6x then
6
dx
dy
10. 10
3. The Power Function Rule
If y = axn
, where a and n are constants
1
n
x
.
a
.
n
dx
dy
i) y = 4x => 4
4 0
x
dx
dy
ii) y = 4x2
=> x
dx
dy
8
iii) y = 4x-2
=>
3
8
x
dx
dy
11. 11
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 3
4
x
dx
dy
(ii) y = 5x + 4 then 5
dx
dy
12. 12
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
du
v
dx
dv
u
dx
dy
13. 13
Examples
If y = u.v dx
du
v
dx
dv
u
dx
dy
i) y = (x+2)(ax2
+bx)
bx
ax
b
ax
x
dx
dy
2
2
2
ii) y = (4x3
-3x+2)(2x2
+4x)
3
12
4
2
4
4
2
3
4 2
2
3
x
x
x
x
x
x
dx
dy
14. 14
6. The Quotient Rule
• If y = u/v where u and v are functions of x
(u = f(x) and v = g(x) ) Then
2
v
dx
dv
u
dx
du
v
dx
dy
15. 15
2
2
4
2
4
1
2
1
4
4
2
x
x
x
x
dx
dy
x
x
y
2
v
dx
dv
u
dx
du
v
dx
dy
then
v
u
y
If
Example 1
16. 16
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
17. 17
Examples
i) y = (ax2
+ bx)½
let v = (ax2
+ bx) , so y = v½
b
ax
.
bx
ax
dx
dy
2
2
1 2
1
2
ii) y = (4x3
+ 3x – 7 )4
let v = (4x3
+ 3x – 7 ), so y = v4
3
12
7
3
4
4 2
3
3
x
.
x
x
dx
dy
dx
dv
.
dv
dy
dx
dy
18. 18
8. The Inverse Function Rule
• Examples
If x = f(y) then
dy
dx
dx
dy 1
i) x = 3y2
then
y
dy
dx
6
so y
dx
dy
6
1
ii) y = 4x3
then
2
12x
dx
dy
so 2
12
1
x
dy
dx
19. 19
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
20. 20
Application I: Marginal Functions
(Revenue, Costs and Profit)
•
Calculating Marginal Functions
dQ
TR
d
MR
dQ
TC
d
MC
21. 21
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
TR
d
MR 6
17
22. 22
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
23. 23
Solution
(i) TC = Q3
– 4Q2
+ 12Q
Then, AC =
TC
/ Q = Q2
– 4Q + 12
(ii) MC =
12
8
3 2
Q
Q
dQ
TC
d
(iii) When does AC = MC?
Q2
– 4Q + 12 = 3Q2
– 8Q + 12
Q = 2
Thus, AC = MC when Q = 2
24. 24
Solution continued….
(iv) When does the slope of AC = 0?
4
2
Q
dQ
AC
d
= 0
Q = 2 when slope AC = 0
(v) Economic Significance?
MC cuts AC curve at minimum point…
25. 25
9. Differentiating Exponential Functions
If y = exp(x) = ex
where e = 2.71828….
then
x
e
dx
dy
More generally,
If y = Aerx
then ry
rAe
dx
dy rx
26. 26
Examples
1) y = e2x
then dx
dy
= 2e2x
2) y = e-7x
then dx
dy
= -7e-7x
27. 27
10. Differentiating Natural Logs
Recall if y = ex
then x = loge y = ln y
If y = ex
then
x
e
dx
dy
= y
From The Inverse Function Rule
y = ex
y
dy
dx 1
Now, if y = ex
this is equivalent to writing
x = ln y
Thus, x = ln y y
dy
dx 1
28. 28
More generally,
if y = ln x x
dx
dy 1
NOTE: the derivative of a natural log
function does not depend on the co-efficient
of x
Thus, if y = ln mx x
dx
dy 1
29. 29
Proof
if y = ln mx m>0
Rules of Logs y = ln m+ ln x
Differentiating (Sum-Difference rule)
x
x
dx
dy 1
1
0
30. 30
Examples
1) y = ln 5x (x>0) x
dx
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
2
2
1
2
1
2
x
.
x
x
dx
dy
1
2
2
2
2
x
x
x
dx
dy
31. 31
3) y = x4
lnx
Product Rule:
3
4
4
1
x
.
x
ln
x
x
dx
dy
= x
ln
x
x 3
3
4
=
x
ln
x 4
1
3
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
4
3
x
x
dx
dy
32. 32
Applications II
• how does demand change with a change in
price……
• ed=
= P
P
Q
Q
= Q
P
.
P
Q
price
in
change
al
proportion
demand
in
change
al
proportion
33. 33
Point elasticity of demand
ed = Q
P
.
dP
dQ
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
34. 34
Example 1
Find ed of the function Q= aP-b
ed = Q
P
.
dP
dQ
ed = b
b
aP
P
.
baP
1
= b
aP
P
.
P
baP
b
b
ed at all price levels is –b
35. 35
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 = Q
P
.
dP
dQ
<0
P is expressed in terms of Q,
1
40
Q
dQ
dP
Inverse rule 40
1
Q
dP
dQ
Hence, ed = Q
P
.
40
1
Q
< 0
Q is 20 ed = 20
78.22
.
40
21
= -2.05
(where P = 200 – 40ln(20+1) = 78.22)
36. 36
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
(ln
d
Take logs and differentiate to find
proportional changes in variables
37. 37
1) Show that if y = x
, then x
y
.
dx
dy
1
and this derivative of ln(y) with respect to x.
Solution:
1
1
1
x
.
y
y
.
dx
dy
= x
x
.
y
1
= x
y
.
.
y
1
= x
38. 38
Solution Continued…
Now ln y = ln x
Re-writing ln y = lnx
x
x
.
dx
)
y
(ln
d
1
Differentiating the ln y with respect to x gives
the proportional change in x.
39. 39
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
changeinp
2
2
1
ct
bt
a
ct
b
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
ct
bt
a
ct
b
dt
)
t
(
P
ln
d
