The Chain Rule allows us to take derivatives of compositions of functions of several variables

1. Lesson 23 (Sections 16.1–2)
The Chain Rule
Math 20
November 14, 2007
Announcements
Problem Set 9 assigned today. Due November 21.
There will be class November 21.
next OH: Today 1-3pm
Midterm II: 12/6, 7-8:30pm in Hall A.

2. 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

3. HW problem 15.7.2
Recall that a discriminating monopolist can choose the
price/quantity sold in two markets.
P1 = a1 − b1 Q1 P2 = a2 − b2 Q2
Suppose cost increases constantly with quantity: C = α(Q1 + Q2 ).
The profit is therefore
π = P1 Q1 + P2 Q2 − α(Q1 + Q2 )
= (a1 − b1 Q1 )Q1 + (a2 − b2 Q2 )Q2 − α(Q1 + Q2 )
2 2
= (a1 − α)Q1 − b1 Q1 + (a2 − α)Q2 − b2 Q2

4. Solution
Completing the square gives
2 2
π = (a1 − α)Q1 − b1 Q1 + (a2 − α)Q2 − b2 Q2
2
(a1 − α) (a1 − α)2
= −b1 Q1 − +
2b1 4b1
2
(a2 − α) (a2 − α)2
− b2 Q 2 − +
2b2 4b2
The optimal quantities are
∗ a1 − α ∗ a2 − α
Q1 = Q2 =
2b1 2b2

5. The corresponding prices are
∗ a1 + α ∗ a2 + α
P1,d = P2,d =
2 2
The maximum profit is
∗ (a1 − α)2 (a2 − α)2
πd = +
4b1 4b2

6. The Indiscriminating Monopolist
An indiscriminating monopolist can only set one price each each
“area”: So
P = a1 − b1 = a2 − b2 Q2 .
This means we can get Q1 and Q2 , and therefore π, all in terms of
P:
a1 − P a2 − P
Q1 = Q2 =
b1 b2
π(P) = P(Q1 + Q2 ) − α(Q1 + Q2 )
a1 − P a2 − P a1 − P a2 − P
=P + −α +
b1 b2 b1 b2
a1 + α a2 + α 1 1 αa1 αa2
= + P− + P2 − −
b1 b2 b1 b2 b1 b2

7. Complete the square in P now and we get
a1 +α a2 +α
b1 + b2 b2 b1
Pi∗ = = ∗
P1,d + P∗
2 1
+ 1 b1 + b2 b1 + b2 2,d
b1 b2
2
a1 +α a2 +α
b1 + b2 αa1 αa2
πi∗ = − −
4 1
+ 1 b1 b2
b1 b2

8. Complete the square in P now and we get
a1 +α a2 +α
b1 + b2 b2 b1
Pi∗ = = ∗
P1,d + P∗
2 1
+ 1 b1 + b2 b1 + b2 2,d
b1 b2
2
a1 +α a2 +α
b1 + b2 αa1 αa2
πi∗ = − −
4 1
+ 1 b1 b2
b1 b2
Subtract and do some algebra:
(a1 − a2 ) 2
πd − πi∗ =
∗
>0
4 (b1 + b2 )

9. 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

10. Optimization of functions of several variables
Last week: algebraic optimization
Critical values of quadratic forms
This week: more differentiation
Chain Rule
Implicit differentiation
Next week: unconstrained optimization
Approximate functions “to second order” and use rules for
quadratic forms
Following week: constrained optimization

11. 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

12. The Chain Rule in one variable
See Section 5.2 for more
(f ◦ g ) (x) = f (g (x)) · g (x)

13. The Chain Rule in one variable
See Section 5.2 for more
(f ◦ g ) (x) = f (g (x)) · g (x)
Or, if y = f (u) and u = f (x), then
dy dy du
=
dx du dx

14. The Chain Rule in one variable
See Section 5.2 for more
(f ◦ g ) (x) = f (g (x)) · g (x)
Or, if y = f (u) and u = f (x), then
dy dy du
=
dx du dx
A goal for today is the chain rule in several variables.

15. Example
Let z = xy 2 , and suppose x and y are given as functions of t:
x = t3 y = sin t
dz
Find dt .

16. Example
Let z = xy 2 , and suppose x and y are given as functions of t:
x = t3 y = sin t
dz
Find dt .
Solution
z = t 3 sin2 t, so
dz
= 3t 2 sin2 t + t 3 2 sin t cos t
dt
dx y2 x dy
dt dt

17. Fact (The Chain Rule, version I)
When z = F (x, y ) with x = f (t) and y = g (t), then
z (t) = F1 (f (t), g (t))f (t) + F2 (f (t), g (t))g (t)
or
dz ∂F dx ∂F dy
= +
dt ∂x dt ∂y dt

18. Fact (The Chain Rule, version I)
When z = F (x, y ) with x = f (t) and y = g (t), then
z (t) = F1 (f (t), g (t))f (t) + F2 (f (t), g (t))g (t)
or
dz ∂F dx ∂F dy
= +
dt ∂x dt ∂y dt
We can generalize to more variables, too. If F is a function of
x1 , x2 , . . . , xn , and each xi is a function of t, then
dz ∂F dx1 ∂F dx2 ∂F dxn
= + + ··· +
dt ∂x1 dt ∂x2 dt ∂xn dt

19. Tree Diagrams for the Chain Rule
F
∂F ∂F
∂x ∂y
x y
dx dy
dt dt
t t
To differentiate with respect to t, find all “leaves” marked t.
Going down each branch, chain (multiply) all the derivatives
together. Then add up the result from each branch.
dz dF ∂F dx ∂F dy
= = +
dt dt ∂x dt ∂y dt

20. Example
Consider a Cobb-Douglas production function defined by
P(A, x, y ) = AK a Lb
where K is capital, L is labor, and A is the “technology” to convert
these quantities to production. Suppose that all of these are
changing over time. Show that
1 dP 1 dA 1 dK 1 dL
= +a +b
P dt A dt K dt L dt
That is
relative rate relative rate relative rate relative rate
of growth in = of growth of + a × of growth of + b × of growth of
output technology capital labor

21. Solution
dP ∂P dA ∂P dK ∂P dL
= + +
dt ∂A dt ∂K dt ∂L dt
dA dK dL
= K a Lb + AaK a−1 Lb + AK a bLb−1
dt dt dt
So
1 dP K a Lb dA + AaK a−1 Lb dK + AK a bLb−1 dL
dt dt dt
=
P dt AK a Lb
1 dA 1 dK 1 dL
= +a +b
A dt K dt L dt

22. 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

23. Fact (The Chain Rule, Version II)
When z = F (x, y ) with x = f (t, s) and y = g (t, s), then
∂z ∂F ∂x ∂F ∂y
= +
∂t ∂x ∂t ∂y ∂t
∂z ∂F ∂x ∂F ∂y
= +
∂s ∂x ∂s ∂y ∂s
F
x y
t s t s

24. Example
∂z ∂z
Suppose z = xy 2 , x = t + s and y = t − s. Find ∂t and ∂s at
(t, z) = (1/2, 1) in two ways:
(i) By expressing z directly in terms of t and s before
differentiating.
(ii) By using the chain rule.

25. Theorem (The Chain Rule, General Version)
Suppose that u is a differentiable function of the n variables
x1 , x2 , . . . , xn , and each xi is a differentiable function of the m
variables t1 , t2 , . . . , tm . Then u is a function of t1 , t2 , . . . , tm and
∂u ∂u ∂x1 ∂u ∂x2 ∂u ∂xn
= + + ··· +
∂ti ∂x1 ∂ti ∂x2 ∂ti ∂xn ∂ti

26. Theorem (The Chain Rule, General Version)
Suppose that u is a differentiable function of the n variables
x1 , x2 , . . . , xn , and each xi is a differentiable function of the m
variables t1 , t2 , . . . , tm . Then u is a function of t1 , t2 , . . . , tm and
∂u ∂u ∂x1 ∂u ∂x2 ∂u ∂xn
= + + ··· +
∂ti ∂x1 ∂ti ∂x2 ∂ti ∂xn ∂ti
In summation notation
n
∂u ∂u ∂xj
=
∂ti ∂xj ∂ti
j=1

27. Example
Write out the Chain Rule for the case where w = f (x, y , z, t) and
x = x(u, v ), y = y (u, v ), z = z(u, v ), and t = t(u, v ).

28. Example
Write out the Chain Rule for the case where w = f (x, y , z, t) and
x = x(u, v ), y = y (u, v ), z = z(u, v ), and t = t(u, v ).
w
x y z t
u v u v u v u v

29. Example
Write out the Chain Rule for the case where w = f (x, y , z, t) and
x = x(u, v ), y = y (u, v ), z = z(u, v ), and t = t(u, v ).
w
x y z t
u v u v u v u v
Solution
∂w ∂w ∂x ∂w ∂y ∂w ∂z ∂w ∂t
= + + +
∂u ∂x ∂u ∂y ∂u ∂z ∂u ∂t ∂u
∂w ∂w ∂x ∂w ∂y ∂w ∂z ∂w ∂t
= + + +
∂v ∂x ∂v ∂y ∂v ∂z ∂v ∂t ∂v

30. 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

31. Matrix Perspective
∂w ∂w ∂x ∂w ∂y ∂w ∂z ∂w ∂t
= + + +
∂u ∂x ∂u ∂y ∂u ∂z ∂u ∂t ∂u
∂w ∂w ∂x ∂w ∂y ∂w ∂z ∂w ∂t
= + + +
∂v ∂x ∂v ∂y ∂v ∂z ∂v ∂t ∂v
Or,
 ∂x ∂x 
 ∂u ∂v 
 ∂y ∂y 
∂w ∂w ∂w ∂w ∂w ∂w 
 ∂u

∂v 
=  ∂z ∂z 
∂u ∂v ∂x ∂y ∂z ∂t  
 ∂u ∂v 
 
∂t ∂t
∂u ∂v

32. 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

33. Leibniz’s Formula for Integrals
Fact
Suppose that f (, t, x), a(t), and b(t) are differentiable functions,
and let
b(t)
F (t) = f (t, x) dx
a(t)

34. Leibniz’s Formula for Integrals
Fact
Suppose that f (, t, x), a(t), and b(t) are differentiable functions,
and let
b(t)
F (t) = f (t, x) dx
a(t)
Then
b(t)
∂f (t, x)
F (t) = f (t, b(t))b (t) − f (t, a(t))a (t) + dx
a(t) ∂t

35. Leibniz’s Formula for Integrals
Fact
Suppose that f (, t, x), a(t), and b(t) are differentiable functions,
and let
b(t)
F (t) = f (t, x) dx
a(t)
Then
b(t)
∂f (t, x)
F (t) = f (t, b(t))b (t) − f (t, a(t))a (t) + dx
a(t) ∂t
Proof.
Apply the chain rule to the function
v
H(t, u, v ) = f (t, x) dx
u
with u = a(t) and v = b(t).

36. Tree Diagram
H
t u v
t t

37. More about the proof
v
H(t, u, v ) = f (t, x) dx
u

38. More about the proof
v
H(t, u, v ) = f (t, x) dx
u
Then by the Fundamental Theorem of Calculus (see Section 10.1)
∂H ∂H
= f (t, v ) = −f (t, u)
∂v ∂u

39. More about the proof
v
H(t, u, v ) = f (t, x) dx
u
Then by the Fundamental Theorem of Calculus (see Section 10.1)
∂H ∂H
= f (t, v ) = −f (t, u)
∂v ∂u
Also,
v
∂H ∂f
= (t, x) dx
∂t u ∂x
since t and x are independent variables.

40. Since F (t) = H(t, a(t), b(t)),
dF ∂H ∂H du ∂H dv
= + +
dt ∂t ∂u dt ∂v dt
b(t)
∂f
= (t, x) + f (t, b(t))b (t) − f (t, a(t))a (t)
a(t) ∂x

41. Application
Example (Example 16.8 with better notation)
Let the profit of a firm be π(t). The present value of the future
profit 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 present
value of all future profit is
T
V (t) = π(τ )e −r (τ −t) dt.
t
Find V (t).

42. Application
Example (Example 16.8 with better notation)
Let the profit of a firm be π(t). The present value of the future
profit 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 present
value of all future profit is
T
V (t) = π(τ )e −r (τ −t) dt.
t
Find V (t).
Answer.
V (t) = rV (t) − π(t)

43. Solution
Since the upper limit is a constant, the only boundary term comes
from the lower limit:
T
∂
V (t) = −π(t)e −r (t−t) + π(τ )e −r τ e rt dτ
t ∂t
T
= −π(t) + r π(τ )e −r τ e rt dτ
t
= rV (t) − π(t).
This means that
π(t) + V (t)
r=
V (t)
So if the fraction on the right is less than the rate of return for
another, “safer” investment like bonds, it would be worth more to
sell the business and buy the bonds.