Download Topic 6: Differentiation

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
1
∆y
∆x
=
2
y 2 − y1
x 2 − x1
3
X
gives
4
slope
5
of
6
the
line
connecting 2 points (x 1 , y 1 ) and (x 2 ,y 2 ) 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
which is different for different values of x
4
Example:A firms cost function is
Y = X2
X
0
1
2
3
4
∆X
Y
∆Y
+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
∆ 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
= lim
f ' ( x) =
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
dx
If y = u .v
du
dv
= u
+ v
dx
dx
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
dv
du
−u
v
dy
dx
dx
=
2
dx
v
14
u
If y =
v
then
dy
=
dx
v
dv
du
−u
dx
dx
v2
Example 1
y =
(x + 2 )
(x + 4 )
(
−2
dy
x + 4 )(1 ) − ( x + 2 )(1 )
=
=
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
dy
dy dv
.
=
dx
dv dx
½
y = (ax + bx)
let v = (ax2 + bx) , so y = v½
1
−
dy
1
2
=
ax + bx 2 .(2 ax + b )
dx
2
(
)
3
ii) y = (4x + 3x – 7 )
4
3
let v = (4x + 3x – 7 ), so y = v
(
dy
3
= 4 4x + 3x − 7
dx
) .(12 x
3
4
2
+3
)
17
8. The Inverse Function Rule
dy
1
=
If x = f(y) then dx dx
dy
• Examples
i)
x = 3y2 then
dx
= 6y
dy
dy
1
so dx = 6 y
ii)
3
y = 4x
dy
= 12 x 2
dx
then
dx
1
so dy = 12 x 2
18
Differentiation in Economics
Application I
•
•
•
•
•
Total Costs = TC = FC + VC
Total Revenue = TR = P * Q
π = Profit = TR – TC
Break even: π = 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
/ Q = Q – 4Q + 12
Then, AC =
d (TC )
2
=
3
Q
− 8 Q + 12
(ii) MC = dQ
(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 + 2x +1
(
2x + 2)
dy
=
dx
x2 + 2x + 1
(
)
30
3) y = x4lnx
Product Rule: ⇒
dy
1
= x 4 + ln x . 4 x 3
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……
proportion al change in demand
proportion al change in price
• ed=
=
∆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
ƒ Price elasticity of demand: ed =
dQ P
.
dP Q
<0
ƒ P is expressed in terms of Q,
dP
40
= −
dQ
Q +1
ƒ Inverse rule ⇒
ƒ Hence, ed =
−
dQ
Q +1
= −
40
dP
Q+ 1 P
.
40 Q
<0
21 78.22
ƒ 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
= dx . y
d (ln y )
=
dx
Take logs and differentiate to
proportional changes in variables
find
36
dy 1
α
1) Show that if y = x , then dx . y = x
α
and this ≡ derivative of ln(y) with respect to x.
Solution:
dy 1
1
.
=
.α x
dx y
y
α −1
xα
1
= y .α x
y
1
= y .α . x
α
= x
37
Solution Continued…
Now ln y = ln x
α
Re-writing ⇒ ln y = αlnx
1
α
d (ln y )
= α. =
⇒
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+ct2
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 + ct2
39