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
y2 y1
b = y =
x x2 x1
2
If the function is non-linear:
40
e.g. if y = x2
30
y=x2
20
10
0
0 1 2 3
X 4 5 6
y y2 y1
x
= x2 x1
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 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
25
y=x2
20
15
10
0
1 2 3 4 5 6 7
4
Example:A firms cost function is
Y = X2
X X Y Y
0 0
1 +1 1 +1
2 +1 4 +3
3 +1 9 +5
4 +1 16 +7
2
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 x0 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
dy
e.g. y = 10 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
i) y = 4x => dx 4 x 0
4
dy
ii) y = 4x 2
=> dx 8 x
dy 3
-2
iii) y = 4x => dx 8 x
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) terms
separately, then add/subtract
dy
(i) y = 2x2 + 3x then 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
If y = u.v u v
dx dx dx
2
i) y = (x+2)(ax +bx)
dy
dx
x 2 2 ax b ax bx
2
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
du dv
v u
u dy dx dx
If y then
v dx v2
Example 1
y
x 2
x 4
dy x 4 1 x 2 1 2
dx x 4 2
x 42
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
dy dy dv
.
Examples dx dv dx
2 ½
i) y = (ax + bx)
let v = (ax2 + bx) , so y = v½
1
dy 1
ax bx 2 .2ax b
2
dx 2
ii) y = (4x + 3x – 7 )
3 4
let v = (4x + 3x – 7 ), so y = v
3 4
dy
dx
3 3
4 4 x 3x 7 . 12 x 3
2
17
8. The Inverse Function Rule
dy 1
If x = f(y) then dx dx
dy
• Examples
i) x = 3y2 then
dx dy 1
dy
6y so dx 6 y
ii) y = 4x3 then
dy dx 1
12 x 2 so dy 12 x 2
dx 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
Solution:
demand curve P=17-3Q
• (i) Find an expression
TR = P.Q = 17Q – 3Q2
for TR in terms of Q
• (ii) Find an expression
for MR in terms of Q d TR
MR 17 6Q
dQ
21
Example 2
22
Solution
(i) TC = Q3 – 4Q2 + 12Q
TC
Then, AC = / Q = Q – 4Q + 12
2
d TC
(ii) MC = dQ 3Q 2
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
x
If y = exp(x) = e where e = 2.71828….
dy
then dx e
x
More generally,
rx
If y = Ae
dy
then dx rAe ry
rx
25
Examples
2x
dy 2x
1) y = e then dx = 2e
-7x
dy -7x
2) y = e then dx = -7e
26
10. Differentiating Natural Logs
Recall if y = ex then x = loge y = ln y
dy
If y = e
x
then ex = y
dx
From The Inverse Function Rule
dx 1
y = e dy y
x
Now, if y = ex this is equivalent to writing
x = ln y
dx 1
Thus, x = ln y 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
dy 1
1) y = ln 5x (x>0)
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
2 x 2
dx
x 2x 1
30
3) y = x4lnx
Product Rule:
dy 4 1
x ln x.4 x3
dx x
= x 3
4 x 3
ln x = x 3
1 4 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……
• e d= proportional change in demand
proportional change in price
Q P Q 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
-b
Find ed of the function Q= aP
dQ P
ed = .
dP Q
b 1 P
ed = baP . 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
dQ P
Price elasticity of demand: ed = .
dP Q
<0
P is expressed in terms of Q,
dP 40
dQ Q 1
dQ Q 1
Inverse rule dP
40
Q 1 P
Hence, ed = 40 . Q <0
21 78.22
Q is 20 ed = 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 d (ln y )
= dx . y = dx
Take logs and differentiate to find
proportional changes in variables
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
lnP(t) = ln(a+bt+ct )
1 dP( t ) b 2ct
. where v = (a+bt+ct2) so lnP = ln v
P( t ) dt a bt ct 2
Using chain rule,
d ln P( t ) b 2ct
dt a bt ct 2
39
