Lesson 23: The Chain Rule

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



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



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.


(f ◦ g)′(x) = f ′(g(x)) · g ′(x)


dy

dx=

dy

du

du

dx



(f ◦ g)′(x) = f ′(g(x)) · g ′(x)


dy

dx=

dy

du

du

dx


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



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


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


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 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




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).



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


H(t, u, v) =

∫ v

uf (t, x) dx




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


H(t, u, v) =

∫ v

uf (t, x) dx


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.


H(t, u, v) =

∫ v

uf (t, x) dx


∂H

∂v= f (t, v)

∂H

∂u= −f (t, u)

Also,∂H

∂t=

∫ v

u

∂f

∂x(t, x) dx



H(t, u, v) =

∫ v

uf (t, x) dx


∂H

∂v= f (t, v)

∂H

∂u= −f (t, u)

Also,∂H

∂t=

∫ v

u

∂f

∂x(t, x) dx


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.

Technology

Lesson 23: The Chain Rule