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Topic 6: Differentiation
Jacques Text Book (edition 4 ):
Chapter 4
1.Rules of Differentiation
2.Applications
2
Differentiation is all about measuring change! Measuring change in a linear function:
y = a + bxa = intercept
b = constant slope i.e. the impact of a unit change in x on the level of y
b = = x
y
12
12xx
yy
3
If the function is non-linear: e.g. if y = x2
0
10
20
30
40
0 1 2 3 4 5 6X
y=
x2
x
y
= 12
12xx
yy
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
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
which is different for different values of x
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
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
dyxf
x
0
lim)('
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
8
Rules for Differentiation (section 4.3)
1. The Constant Rule
If y = c where c is a constant,
0dx
dy
e.g. y = 10 then 0dx
dy
9
2. The Linear Function Rule
If y = a + bx
bdx
dy
e.g. y = 10 + 6x then 6dx
dy
10
3. The Power Function Rule
If y = axn, where a and n are constants
1 nx.a.ndx
dy
i) y = 4x => 44 0 xdx
dy
ii) y = 4x2 => xdx
dy8
iii) y = 4x-2 => 38 x
dx
dy
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 34 xdx
dy
(ii) y = 5x + 4 then 5dx
dy
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
duv
dx
dvu
dx
dy
13
Examples
If y = u.v dx
duv
dx
dvu
dx
dy
i ) y = ( x + 2 ) ( a x 2 + b x )
bxaxbaxxdxdy 222
i i ) y = ( 4 x 3 - 3 x + 2 ) ( 2 x 2 + 4 x )
3124244234 223 xxxxxx
dxdy
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
2vdxdv
udxdu
v
dx
dy
15
22 4
2
4
1214
4
2
xx
xx
dx
dy
x
xy
2vdxdv
udxdu
v
dx
dythen
v
uyIf
Example 1
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
Examplesi) y = (ax2 + bx)½
let v = (ax2 + bx) , so y = v½
bax.bxaxdx
dy
2
2
121
2
ii) y = (4x3 + 3x – 7 )4
let v = (4x3 + 3x – 7 ), so y = v4
3127344 233 x.xxdx
dy
dx
dv.
dv
dy
dx
dy
18
8. The Inverse Function Rule
• Examples
If x = f(y) then dy
dxdx
dy 1
i) x = 3y2 then
ydy
dx6 so ydx
dy
6
1
ii) y = 4x3 then
212x
dx
dy so 212
1
xdy
dx
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
Application I: Marginal Functions (Revenue, Costs and Profit)
•Calculating Marginal Functions
dQ
TRdMR
dQ
TCdMC
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
TRdMR 617
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
Solution
(i) TC = Q3 – 4Q2 + 12Q
Then, AC = TC / Q = Q2 – 4Q + 12
(ii) MC =
1283 2 QQdQ
TCd
(iii) When does AC = MC? Q2 – 4Q + 12 = 3Q2 – 8Q + 12
Q = 2 Thus, AC = MC when Q = 2
24
Solution continued….
(iv) When does the slope of AC = 0?
42 QdQ
ACd= 0
Q = 2 when slope AC = 0 (v) Economic Significance? MC cuts AC curve at minimum point…
25
9. Differentiating Exponential Functions
If y = exp(x) = ex where e = 2.71828….
then xe
dx
dy
More generally,
If y = Aerx
then ryrAedx
dy rx
26
Examples
1) y = e2x then dx
dy = 2e2x
2) y = e-7x then dx
dy = -7e-7x
27
10. Differentiating Natural Logs
Recall if y = ex then x = loge y = ln y
If y = ex then xedx
dy = y
From The Inverse Function Rule
y = ex ydy
dx 1
Now, if y = ex this is equivalent to writing x = ln y
Thus, x = ln y ydy
dx 1
28
More generally,
if y = ln x xdx
dy 1
NOTE: the derivative of a natural log function does not depend on the co-efficient of x
Thus, if y = ln mx xdx
dy 1
29
Proof
if y = ln mx m>0
Rules of Logs y = ln m+ ln x
Differentiating (Sum-Difference rule)
xxdx
dy 110
30
Examples1) y = ln 5x (x>0)
xdx
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
2212
12
x.
xxdx
dy
12
222
xx
x
dx
dy
31
3) y = x4lnx
Product Rule:
34 41
x.xlnx
xdx
dy
= xlnxx 33 4 = xlnx 413
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
43
xxdx
dy
32
Applications II
• how does demand change with a change in price……
• ed=
= P
P
Q
Q = Q
P.
PQ
priceinchangealproportion
demandinchangealproportion
33
Point elasticity of demand
ed = QP.
dPdQ
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
Example 1
Find ed of the function Q= aP-b
ed = QP.
dPdQ
ed = bb
aP
P.baP
1
= baP
P.
PbaP
b
b
ed at all price levels is –b
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 = QP.
dPdQ <0
P is expressed in terms of Q,
1
40
QdQ
dP
Inverse rule 40
1
Q
dP
dQ
Hence, ed = QP.
401Q < 0
Q is 20 ed = 2078.22.
4021 = -2.05
(where P = 200 – 40ln(20+1) = 78.22)
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(lnd
Take logs and differentiate to find proportional changes in variables
37
1) Show that if y = x, then xy.
dx
dy
1
and this derivative of ln(y) with respect to x.
Solution:
111 x.yy
.dx
dy
= x
x.
y
1
= x
y..
y1
= x
38
Solution Continued…
Now ln y = ln x
Re-writing ln y = lnx
xx.
dx
)y(lnd
1
Differentiating the ln y with respect to x gives the proportional change in x.
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 change in p
2
21
ctbta
ctb
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
ctbta
ctb
dt
)t(Plnd