Topic 6: Differentiation
Jacques Text Book (edition 4 ):
Chapter 4
1.Rules of Differentiation
2.Applications
1
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 = y
x
=
y2  y1
x2  x1
2
40
If the function is non-linear:
e.g. if y = x2
y=x2
30
20
10
0
0
y
x
1
=
2
y 2  y1
x2  x1
3
X
4
5
gives slope of the
6
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
3
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
40
35
30
y=x2
25
20
15
10
5
0
1
2
3
4
5
6
7
X
4
Example:A firms cost function is
Y = X2
X
0
1
2
3
4
2
X
Y
Y
+1
+1
+1
+1
0
1
4
9
16
+1
+3
+5
+7
Y=X
Y+Y = (X+X) 2
Y+Y =X2+2X.X+X2
Y = X2+2X.X+X2 – Y
since Y = X2
 Y =
2X.X+X2
Y
X
= 2X+X
The slope depends on X and X
5
The slope of the graph of a function
is called the derivative of the
function
dy
y
f ' ( x) 
 lim
dx x0 x
• 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
6
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
7
Rules for Differentiation
(section 4.3)
1. The Constant Rule
If y = c
where c is a constant,
dy
0
dx
e.g. y = 10
dy
then dx  0
8
2. The Linear Function Rule
If y = a + bx
dy
b
dx
dy
6
e.g. y = 10 + 6x then dx
9
3. The Power Function Rule
If y = axn,
where a and n are constants
dy
 n.a.x n1
dx
dy
0

4
x
4
i) y = 4x => dx
ii) y = 4x
2
dy
=> dx  8 x
dy
3


8
x
iii) y = 4x => dx
-2
10
4. The Sum-Difference Rule
If y = f(x)  g(x)
dy d [ f ( x )] d [ g( x )]


dx
dx
dx
If y is the sum/difference of two or more
functions of x:
differentiate the 2 (or more)
separately, then add/subtract
(i) y = 2x2 + 3x
then
terms
dy
 4x  3
dx
dy
(ii) y = 5x + 4 then dx  5
11
5. The Product Rule
If y = u.v where u and v are functions of x,
(u = f(x) and v = g(x) ) Then
dy
dv
du
 u
v
dx
dx
dx
12
Examples
dy
dv
du
 u
 v
dx
dx
dx
If y = u.v
2
i) y = (x+2)(ax +bx)
dy
  x  2 2 ax  b   ax 2  bx
dx


ii) y = (4x3-3x+2)(2x2+4x)
dy   4x3 3x  2 4x  4  2x2  4x 12x2 3




dx 
13
6. The Quotient Rule
• If y = u/v where u and v are functions of x
(u = f(x) and v = g(x) )
Then
du
dv
v
u
dy
dx
dx

2
dx
v
14
u
If y 
v
then
dy

dx
v
du
dv
u
dx
dx
v2
Example 1

x  2
y
 x  4

dy
x  4 1   x  2 1
2


2
2
dx
 x  4
 x  4
15
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
dy dy dv

.
dx dv dx
16
Examples
i)
2
y = (ax + bx)
dy dy dv

.
dx dv dx
½
let v = (ax2 + bx) , so y = v½
1

dy 1
2

ax  bx 2 .2ax  b 
dx 2


ii) y = (4x + 3x – 7 )
3
4
let v = (4x + 3x – 7 ), so y = v
3


4

3
dy
3
2
 4 4 x  3x  7 . 12 x  3
dx
17
8. The Inverse Function Rule
dy
1

If x = f(y) then dx dx
dy
• Examples
i)
x = 3y2 then
dx
 6y
dy
ii)
dy
1
so dx  6 y
y = 4x3 then
dy
 12x 2
dx
dx
1
so dy  12 x 2
18
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
19
Application I: Marginal Functions
(Revenue, Costs and Profit)
•
Calculating Marginal Functions
d TR 
MR 
dQ
d TC 
MC 
dQ
20
Example 1
• A firm faces the
demand curve P=173Q
• (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
d TR 
MR 
 17  6Q
dQ
21
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
22
Solution
(i) TC = Q3 – 4Q2 + 12Q
TC
2
Then, AC =
/ Q = Q – 4Q + 12
d TC 
2
(ii) MC = dQ  3Q  8Q  12
(iii) When does AC = MC?
Q2 – 4Q + 12 = 3Q2 – 8Q + 12
Q =2
Thus, AC = MC when Q = 2
23
Solution continued….
(iv) When does the slope of AC = 0?
d  AC 
 2Q  4 = 0
dQ
 Q = 2 when slope AC = 0
(v) Economic Significance?
MC cuts AC curve at minimum point…
24
9. Differentiating Exponential Functions
If y = exp(x) = e
x
where e = 2.71828….
dy
x
then dx  e
More generally,
rx
If y = Ae
dy
rx
then dx  rAe  ry
25
Examples
1) y = e
2x
-7x
2) y = e
dy
2x
then dx = 2e
dy
-7x
then dx = -7e
26
10. Differentiating Natural Logs
Recall if y = ex then x = loge y = ln y
If y = e
x
then
dy
 ex
dx
= y
From The Inverse Function Rule
dx
1
y = e  dy  y
x
Now, if y = ex this is equivalent to writing
x = ln y
Thus, x = ln y
dx
1

 dy y
27
More generally,
dy 1
if y = ln x  dx  x
NOTE: the derivative of a natural log
function does not depend on the co-efficient
of x
Thus, if y = ln mx 
dy 1

dx x
28
Proof
 if y = ln mx
m>0
 Rules of Logs  y = ln m+ ln x
 Differentiating (Sum-Difference rule)
dy
1 1
0 
dx
x x
29
Examples
1) y = ln 5x (x>0) 
dy 1

dx x
2
2) y = ln(x +2x+1)
2
let v = (x +2x+1)
so y = ln v
dy dy dv
Chain Rule:  dx  dv . dx
dy
1
 2
.2 x  2
dx x  2 x  1

dy
2 x  2
 2
dx
x  2x  1


30
3) y = x4lnx
Product Rule: 
dy
4 1
x
 ln x.4 x3
dx
x
3
3
3
1  4 ln x 
x
x

4
x
ln
x
=
=
4) y = ln(x3(x+2)4)
Simplify first using rules of logs
 y = lnx3 + ln(x+2)4
 y = 3lnx + 4ln(x+2)
dy 3
4
 
dx x x  2
31
Applications II
• how does demand change with a change in
price……
proportional change in demand
proportional change in price
• e d=
=
Q
Q
P
P
Q P
= P . Q
32
Point elasticity of demand
dQ P
.
ed = dP Q
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
33
Example 1
Find ed of the function Q= aP
ed =
ed =
-b
dQ P
.
dP Q
 baP
b 1
P
. b
aP
 baP b P
= P . aP b  b
ed at all price levels is –b
34
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 =
dQ P
.
dP Q
<0
 P is expressed in terms of Q,
dP
40

dQ
Q 1
 Inverse rule 
 Hence, ed =
 Q is 20  ed

dQ
Q 1

dP
40
Q 1 P
.
40 Q
<0
21 78.22
=  40 . 20 = -2.05
(where P = 200 – 40ln(20+1) = 78.22)
35
Application III: Differentiation of Natural
Logs to find Proportional Changes
f’(x)
The derivative of log(f(x)) 
/f(x), or the
proportional change in the variable x
i.e. y = f(x), then the proportional  x
dy 1
= dx . y
d (ln y )
=
dx
Take logs and differentiate to
proportional changes in variables
find
36
dy 1

1) Show that if y = x , then dx . y  x

and this  derivative of ln(y) with respect to x.
Solution:
dy 1
1
. 
.x  1
dx y
y
1
x
= y . x
1
y
= y . . x

= x
37
Solution Continued…
Now ln y = ln x

Re-writing  ln y = lnx
d (ln y )
1 
 . 

dx
x
x
Differentiating the ln y with respect to x gives
the proportional change in x.
38
Example 2: If Price level at time t is
P(t) = a+bt+ct2
Calculate the rate of inflation.
Solution:
Alternatively,
The inflation rate at t is the proportional differentiating the log of P(t) wrt t directly
change in p
2
1 dP( t ) b  2ct
.

P( t ) dt a  bt  ct 2
lnP(t) = ln(a+bt+ct )
where v = (a+bt+ct2) so lnP = ln v
Using chain rule,
d ln P( t )
b  2ct

dt
a  bt  ct 2
39