lesson 23: the chain rule
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Lesson 23 (Sections 16.1–2)The Chain Rule
Math 20
November 14, 2007
Announcements
I Problem Set 9 assigned today. Due November 21.
I There will be class November 21.
I next OH: Today 1-3pm
I Midterm II: 12/6, 7-8:30pm in Hall A.
Outline
HW problem 15.7.2
Where we’re going
Chain Rule I
Chain Rule II
Matrix expressions for the Chain Rule
Leibniz’s Formula for Integrals
HW problem 15.7.2
Recall that a discriminating monopolist can choose theprice/quantity sold in two markets.
P1 = a1 − b1Q1 P2 = a2 − b2Q2
Suppose cost increases constantly with quantity: C = α(Q1 + Q2).The profit is therefore
π = P1Q1 + P2Q2 − α(Q1 + Q2)
= (a1 − b1Q1)Q1 + (a2 − b2Q2)Q2 − α(Q1 + Q2)
= (a1 − α)Q1 − b1Q21 + (a2 − α)Q2 − b2Q2
2
Solution
Completing the square gives
π = (a1 − α)Q1 − b1Q21 + (a2 − α)Q2 − b2Q2
2
= −b1
(Q1 −
(a1 − α)
2b1
)2
+(a1 − α)2
4b1
− b2
(Q2 −
(a2 − α)
2b2
)2
+(a2 − α)2
4b2
The optimal quantities are
Q∗1 =a1 − α
2b1Q∗2 =
a2 − α
2b2
The corresponding prices are
P∗1,d =a1 + α
2P∗2,d =
a2 + α
2
The maximum profit is
π∗d =(a1 − α)2
4b1+
(a2 − α)2
4b2
The Indiscriminating Monopolist
An indiscriminating monopolist can only set one price each each“area”: So
P = a1 − b1 = a2 − b2Q2.
This means we can get Q1 and Q2, and therefore π, all in terms ofP:
Q1 =a1 − P
b1Q2 =
a2 − P
b2
π(P) = P(Q1 + Q2) − α(Q1 + Q2)
= P
(a1 − P
b1+
a2 − P
b2
)− α
(a1 − P
b1+
a2 − P
b2
)=
(a1 + α
b1+
a2 + α
b2
)P −
(1
b1+
1
b2
)P2 − αa1
b1− αa2
b2
Complete the square in P now and we get
P∗i =a1+α
b1+ a2+α
b2
2(
1b1
+ 1b2
) =b2
b1 + b2P∗1,d +
b1
b1 + b2P∗2,d
π∗i =
(a1+α
b1+ a2+α
b2
)2
4(
1b1
+ 1b2
) − αa1
b1− αa2
b2
Subtract and do some algebra:
π∗d − π∗i =(a1 − a2) 2
4 (b1 + b2)> 0
Complete the square in P now and we get
P∗i =a1+α
b1+ a2+α
b2
2(
1b1
+ 1b2
) =b2
b1 + b2P∗1,d +
b1
b1 + b2P∗2,d
π∗i =
(a1+α
b1+ a2+α
b2
)2
4(
1b1
+ 1b2
) − αa1
b1− αa2
b2
Subtract and do some algebra:
π∗d − π∗i =(a1 − a2) 2
4 (b1 + b2)> 0
Outline
HW problem 15.7.2
Where we’re going
Chain Rule I
Chain Rule II
Matrix expressions for the Chain Rule
Leibniz’s Formula for Integrals
Optimization of functions of several variables
I Last week: algebraic optimizationI Critical values of quadratic forms
I This week: more differentiationI Chain RuleI Implicit differentiation
I Next week: unconstrained optimizationI Approximate functions “to second order” and use rules for
quadratic forms
I Following week: constrained optimization
Outline
HW problem 15.7.2
Where we’re going
Chain Rule I
Chain Rule II
Matrix expressions for the Chain Rule
Leibniz’s Formula for Integrals
The Chain Rule in one variableSee Section 5.2 for more
(f ◦ g)′(x) = f ′(g(x)) · g ′(x)
Or, if y = f (u) and u = f (x), then
dy
dx=
dy
du
du
dx
A goal for today is the chain rule in several variables.
The Chain Rule in one variableSee Section 5.2 for more
(f ◦ g)′(x) = f ′(g(x)) · g ′(x)
Or, if y = f (u) and u = f (x), then
dy
dx=
dy
du
du
dx
A goal for today is the chain rule in several variables.
The Chain Rule in one variableSee Section 5.2 for more
(f ◦ g)′(x) = f ′(g(x)) · g ′(x)
Or, if y = f (u) and u = f (x), then
dy
dx=
dy
du
du
dx
A goal for today is the chain rule in several variables.
Example
Let z = xy2, and suppose x and y are given as functions of t:
x = t3 y = sin t
Find dzdt .
Solutionz = t3 sin2 t, so
dz
dt= 3t2︸︷︷︸
dxdt
sin2 t︸ ︷︷ ︸y2
+ t3︸︷︷︸x
2 sin t cos t︸ ︷︷ ︸dydt
Example
Let z = xy2, and suppose x and y are given as functions of t:
x = t3 y = sin t
Find dzdt .
Solutionz = t3 sin2 t, so
dz
dt= 3t2︸︷︷︸
dxdt
sin2 t︸ ︷︷ ︸y2
+ t3︸︷︷︸x
2 sin t cos t︸ ︷︷ ︸dydt
Fact (The Chain Rule, version I)
When z = F (x , y) with x = f (t) and y = g(t), then
z ′(t) = F ′1(f (t), g(t))f ′(t) + F ′2(f (t), g(t))g ′(t)
or
dz
dt=∂F
∂x
dx
dt+∂F
∂y
dy
dt
We can generalize to more variables, too. If F is a function ofx1, x2, . . . , xn, and each xi is a function of t, then
dz
dt=∂F
∂x1
dx1
dt+∂F
∂x2
dx2
dt+ · · · +
∂F
∂xn
dxn
dt
Fact (The Chain Rule, version I)
When z = F (x , y) with x = f (t) and y = g(t), then
z ′(t) = F ′1(f (t), g(t))f ′(t) + F ′2(f (t), g(t))g ′(t)
or
dz
dt=∂F
∂x
dx
dt+∂F
∂y
dy
dt
We can generalize to more variables, too. If F is a function ofx1, x2, . . . , xn, and each xi is a function of t, then
dz
dt=∂F
∂x1
dx1
dt+∂F
∂x2
dx2
dt+ · · · +
∂F
∂xn
dxn
dt
Tree Diagrams for the Chain Rule
F
x
t
dxdt
∂F∂x
y
t
dydt
∂F∂y
To differentiate with respect to t, find all “leaves” marked t.Going down each branch, chain (multiply) all the derivativestogether. Then add up the result from each branch.
dz
dt=
dF
dt=∂F
∂x
dx
dt+∂F
∂y
dy
dt
Example
Consider a Cobb-Douglas production function defined by
P(A, x , y) = AK aLb
where K is capital, L is labor, and A is the “technology” to convertthese quantities to production. Suppose that all of these arechanging over time. Show that
1
P
dP
dt=
1
A
dA
dt+ a
1
K
dK
dt+ b
1
L
dL
dt
That is
relative rateof growth in
output=
relative rateof growth oftechnology
+ a ×relative rateof growth of
capital+ b ×
relative rateof growth of
labor
Solution
dP
dt=∂P
∂A
dA
dt+∂P
∂K
dK
dt+∂P
∂L
dL
dt
= K aLb dA
dt+ AaK a−1Lb dK
dt+ AK abLb−1 dL
dt
So
1
P
dP
dt=
K aLb dAdt + AaK a−1Lb dK
dt + AK abLb−1 dLdt
AK aLb
=1
A
dA
dt+ a
1
K
dK
dt+ b
1
L
dL
dt
Outline
HW problem 15.7.2
Where we’re going
Chain Rule I
Chain Rule II
Matrix expressions for the Chain Rule
Leibniz’s Formula for Integrals
Fact (The Chain Rule, Version II)
When z = F (x , y) with x = f (t, s) and y = g(t, s), then
∂z
∂t=∂F
∂x
∂x
∂t+∂F
∂y
∂y
∂t∂z
∂s=∂F
∂x
∂x
∂s+∂F
∂y
∂y
∂s
F
x
t s
y
t s
Example
Suppose z = xy2, x = t + s and y = t − s. Find ∂z∂t and ∂z
∂s at(t, z) = (1/2, 1) in two ways:
(i) By expressing z directly in terms of t and s beforedifferentiating.
(ii) By using the chain rule.
Theorem (The Chain Rule, General Version)
Suppose that u is a differentiable function of the n variablesx1, x2, . . . , xn, and each xi is a differentiable function of the mvariables t1, t2, . . . , tm. Then u is a function of t1, t2, . . . , tm and
∂u
∂ti=
∂u
∂x1
∂x1
∂ti+∂u
∂x2
∂x2
∂ti+ · · · +
∂u
∂xn
∂xn
∂ti
In summation notation
∂u
∂ti=
n∑j=1
∂u
∂xj
∂xj
∂ti
Theorem (The Chain Rule, General Version)
Suppose that u is a differentiable function of the n variablesx1, x2, . . . , xn, and each xi is a differentiable function of the mvariables t1, t2, . . . , tm. Then u is a function of t1, t2, . . . , tm and
∂u
∂ti=
∂u
∂x1
∂x1
∂ti+∂u
∂x2
∂x2
∂ti+ · · · +
∂u
∂xn
∂xn
∂ti
In summation notation
∂u
∂ti=
n∑j=1
∂u
∂xj
∂xj
∂ti
Example
Write out the Chain Rule for the case where w = f (x , y , z , t) andx = x(u, v), y = y(u, v), z = z(u, v), and t = t(u, v).
w
x
u v
y
u v
z
u v
t
u v
Solution
∂w
∂u=∂w
∂x
∂x
∂u+∂w
∂y
∂y
∂u+∂w
∂z
∂z
∂u+∂w
∂t
∂t
∂u∂w
∂v=∂w
∂x
∂x
∂v+∂w
∂y
∂y
∂v+∂w
∂z
∂z
∂v+∂w
∂t
∂t
∂v
Example
Write out the Chain Rule for the case where w = f (x , y , z , t) andx = x(u, v), y = y(u, v), z = z(u, v), and t = t(u, v).
w
x
u v
y
u v
z
u v
t
u v
Solution
∂w
∂u=∂w
∂x
∂x
∂u+∂w
∂y
∂y
∂u+∂w
∂z
∂z
∂u+∂w
∂t
∂t
∂u∂w
∂v=∂w
∂x
∂x
∂v+∂w
∂y
∂y
∂v+∂w
∂z
∂z
∂v+∂w
∂t
∂t
∂v
Example
Write out the Chain Rule for the case where w = f (x , y , z , t) andx = x(u, v), y = y(u, v), z = z(u, v), and t = t(u, v).
w
x
u v
y
u v
z
u v
t
u v
Solution
∂w
∂u=∂w
∂x
∂x
∂u+∂w
∂y
∂y
∂u+∂w
∂z
∂z
∂u+∂w
∂t
∂t
∂u∂w
∂v=∂w
∂x
∂x
∂v+∂w
∂y
∂y
∂v+∂w
∂z
∂z
∂v+∂w
∂t
∂t
∂v
Outline
HW problem 15.7.2
Where we’re going
Chain Rule I
Chain Rule II
Matrix expressions for the Chain Rule
Leibniz’s Formula for Integrals
Matrix Perspective
∂w
∂u=∂w
∂x
∂x
∂u+∂w
∂y
∂y
∂u+∂w
∂z
∂z
∂u+∂w
∂t
∂t
∂u∂w
∂v=∂w
∂x
∂x
∂v+∂w
∂y
∂y
∂v+∂w
∂z
∂z
∂v+∂w
∂t
∂t
∂v
Or,
(∂w
∂u
∂w
∂v
)=
(∂w
∂x
∂w
∂y
∂w
∂z
∂w
∂t
)
∂x
∂u
∂x
∂v∂y
∂u
∂y
∂v∂z
∂u
∂z
∂v∂t
∂u
∂t
∂v
Outline
HW problem 15.7.2
Where we’re going
Chain Rule I
Chain Rule II
Matrix expressions for the Chain Rule
Leibniz’s Formula for Integrals
Leibniz’s Formula for Integrals
FactSuppose that f (, t, x), a(t), and b(t) are differentiable functions,and let
F (t) =
∫ b(t)
a(t)f (t, x) dx
Then
F ′(t) = f (t, b(t))b′(t) − f (t, a(t))a′(t) +
∫ b(t)
a(t)
∂f (t, x)
∂tdx
Proof.Apply the chain rule to the function
H(t, u, v) =
∫ v
uf (t, x) dx
with u = a(t) and v = b(t).
Leibniz’s Formula for Integrals
FactSuppose that f (, t, x), a(t), and b(t) are differentiable functions,and let
F (t) =
∫ b(t)
a(t)f (t, x) dx
Then
F ′(t) = f (t, b(t))b′(t) − f (t, a(t))a′(t) +
∫ b(t)
a(t)
∂f (t, x)
∂tdx
Proof.Apply the chain rule to the function
H(t, u, v) =
∫ v
uf (t, x) dx
with u = a(t) and v = b(t).
Leibniz’s Formula for Integrals
FactSuppose that f (, t, x), a(t), and b(t) are differentiable functions,and let
F (t) =
∫ b(t)
a(t)f (t, x) dx
Then
F ′(t) = f (t, b(t))b′(t) − f (t, a(t))a′(t) +
∫ b(t)
a(t)
∂f (t, x)
∂tdx
Proof.Apply the chain rule to the function
H(t, u, v) =
∫ v
uf (t, x) dx
with u = a(t) and v = b(t).
Tree Diagram
H
t u
t
v
t
More about the proof
H(t, u, v) =
∫ v
uf (t, x) dx
Then by the Fundamental Theorem of Calculus (see Section 10.1)
∂H
∂v= f (t, v)
∂H
∂u= −f (t, u)
Also,∂H
∂t=
∫ v
u
∂f
∂x(t, x) dx
since t and x are independent variables.
More about the proof
H(t, u, v) =
∫ v
uf (t, x) dx
Then by the Fundamental Theorem of Calculus (see Section 10.1)
∂H
∂v= f (t, v)
∂H
∂u= −f (t, u)
Also,∂H
∂t=
∫ v
u
∂f
∂x(t, x) dx
since t and x are independent variables.
More about the proof
H(t, u, v) =
∫ v
uf (t, x) dx
Then by the Fundamental Theorem of Calculus (see Section 10.1)
∂H
∂v= f (t, v)
∂H
∂u= −f (t, u)
Also,∂H
∂t=
∫ v
u
∂f
∂x(t, x) dx
since t and x are independent variables.
Since F (t) = H(t, a(t), b(t)),
dF
dt=∂H
∂t+∂H
∂u
du
dt+∂H
∂v
dv
dt
=
∫ b(t)
a(t)
∂f
∂x(t, x) + f (t, b(t))b′(t) − f (t, a(t))a′(t)
Application
Example (Example 16.8 with better notation)
Let the profit of a firm be π(t). The present value of the futureprofit π(τ) where τ > t is
π(τ)e−r(τ−t),
where r is the discount rate. On a time interval [0,T ], the presentvalue of all future profit is
V (t) =
∫ T
tπ(τ)e−r(τ−t) dt.
Find V ′(t).
Answer.
V ′(t) = rV (t) − π(t)
Application
Example (Example 16.8 with better notation)
Let the profit of a firm be π(t). The present value of the futureprofit π(τ) where τ > t is
π(τ)e−r(τ−t),
where r is the discount rate. On a time interval [0,T ], the presentvalue of all future profit is
V (t) =
∫ T
tπ(τ)e−r(τ−t) dt.
Find V ′(t).
Answer.
V ′(t) = rV (t) − π(t)
SolutionSince the upper limit is a constant, the only boundary term comesfrom the lower limit:
V ′(t) = −π(t)e−r(t−t) +
∫ T
t
∂
∂tπ(τ)e−rτert dτ
= −π(t) + r
∫ T
tπ(τ)e−rτert dτ
= rV (t) − π(t).
This means that
r =π(t) + V ′(t)
V (t)
So if the fraction on the right is less than the rate of return foranother, “safer” investment like bonds, it would be worth more tosell the business and buy the bonds.
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