Lesson 24 (Sections 16.3, 16.7)Implicit Differentiation
Math 20
November 16, 2007
Announcements
I Problem Set 9 on the website. Due November 21.
I There will be class November 21 and homework dueNovember 28.
I next OH: Monday 1-2pm, Tuesday 3-4pm
I Midterm II: Thursday, 12/6, 7-8:30pm in Hall A.
I Go Harvard! Beat Yale!
Outline
Cleanup on Leibniz Rule
Implicit Differentiation in two dimensionsThe Math 1a wayOld school Implicit DifferentationNew school Implicit DifferentationCompare
Application
More than two dimensions
The second derivative
Last Time: 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
Last Time: 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
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.
Outline
Cleanup on Leibniz Rule
Implicit Differentiation in two dimensionsThe Math 1a wayOld school Implicit DifferentationNew school Implicit DifferentationCompare
Application
More than two dimensions
The second derivative
An example
Consider theutility function
u(x , y) = −1
x−1
y
What is theslope of thetangent linealong theindifferencecurveu(x , y) = −1?
1 2 3 4
1
2
3
4
The Math 1a way
Solve for y in terms of x and differentiate:
1
x+
1
y= 1 =⇒ y =
1
1− 1/x
So
dy
dx=
−1
(1− 1/x)2
(1
x2
)=
−1
x2(1− 1/x)2=
−1
(x − 1)2
Old school Implicit Differentation
Differentiate the equation remembering that y is presumed to be afunction of x :
− 1
x2− 1
y2
dy
dx= 0
Sody
dx= −y2
x2= −
(y
x
)2
New school Implicit Differentation
This is a formalized version of old school: If
F (x , y) = c
Then by differentiating the equation and treating y as a functionof x , we get
∂F
∂x+∂F
∂y
(dy
dx
)F
= 0
So (dy
dx
)F
= −∂F/∂x
∂F/∂x
The (·)F notation reminds us that y is not explicitly a function ofx , but if F is held constant we can treat it implicitly so.
Tree diagram
F
x y
x
∂F
∂x+∂F
∂y
(dy
dx
)F
= 0
The big idea
FactAlong the level curve F (x , y) = c, the slope of the tangent line isgiven by
dy
dx=
(dy
dx
)F
= −∂F/∂x
∂F/∂x= −F ′1(x , y)
F ′2(x , y)
Compare
I Explicitly solving for y is tedious, and sometimes impossible.
I Either implicit method brings out more clearly the important
fact that (in our example)(
dydx
)u
depends only on the ratio
y/x.
I Old-school implicit differentiation is familiar but (IMO)contrived.
I New-school implicit differentiation is systematic andgeneralizable.
Outline
Cleanup on Leibniz Rule
Implicit Differentiation in two dimensionsThe Math 1a wayOld school Implicit DifferentationNew school Implicit DifferentationCompare
Application
More than two dimensions
The second derivative
Application
If u(x , y) is a utility function of two goods, then u(x , y) = c is aindifference curve, and the slope represents the marginal rate ofsubstitution:
Ryx = −(
dy
dx
)u
=u′xu′y
=MUx
MUy
Outline
Cleanup on Leibniz Rule
Implicit Differentiation in two dimensionsThe Math 1a wayOld school Implicit DifferentationNew school Implicit DifferentationCompare
Application
More than two dimensions
The second derivative
More than two dimensions
The basic idea is to close your eyes and use the chain rule:
Example
Suppose a surface is given by F (x , y , z) = c . If this defines z as afunction of x and y , find z ′x and z ′y .
SolutionSetting F (x , y , z) = c and remembering z is implicitly a functionof x and y, we get
∂F
∂x+∂F
∂z
(∂z
∂x
)F
= 0 =⇒(∂z
∂x
)F
= −F ′xF ′z
∂F
∂y+∂F
∂z
(∂z
∂y
)F
= 0 =⇒(∂z
∂y
)F
= −F ′yF ′z
More than two dimensions
The basic idea is to close your eyes and use the chain rule:
Example
Suppose a surface is given by F (x , y , z) = c . If this defines z as afunction of x and y , find z ′x and z ′y .
SolutionSetting F (x , y , z) = c and remembering z is implicitly a functionof x and y, we get
∂F
∂x+∂F
∂z
(∂z
∂x
)F
= 0 =⇒(∂z
∂x
)F
= −F ′xF ′z
∂F
∂y+∂F
∂z
(∂z
∂y
)F
= 0 =⇒(∂z
∂y
)F
= −F ′yF ′z
Tree diagram
F
x y z
x
∂F
∂x+∂F
∂z
(∂z
∂x
)F
= 0 =⇒(∂z
∂x
)F
= −F ′xF ′z
Example
Suppose production is given by a Cobb-Douglas function
P(A,K , L) = AK aLb
where K is capital, L is labor, and A is technology. Compute thechanges in technology or capital needed to sustain production iflabor decreases.
Solution
(∂K
∂L
)P
= −P ′LP ′K
= −AK abLb−1
AaK a−1Lb= −b
a· K
L(∂A
∂L
)P
= −P ′LP ′A
= −AK abLb−1
K aLb= −bA
L
So if labor decreases by 1 unit we need either ba ·
KL more capital or
bAL more tech to sustain production.
Example
Suppose production is given by a Cobb-Douglas function
P(A,K , L) = AK aLb
where K is capital, L is labor, and A is technology. Compute thechanges in technology or capital needed to sustain production iflabor decreases.
Solution
(∂K
∂L
)P
= −P ′LP ′K
= −AK abLb−1
AaK a−1Lb= −b
a· K
L(∂A
∂L
)P
= −P ′LP ′A
= −AK abLb−1
K aLb= −bA
L
So if labor decreases by 1 unit we need either ba ·
KL more capital or
bAL more tech to sustain production.
Outline
Cleanup on Leibniz Rule
Implicit Differentiation in two dimensionsThe Math 1a wayOld school Implicit DifferentationNew school Implicit DifferentationCompare
Application
More than two dimensions
The second derivative
The second derivative: Derivation
What is the concavity of an indifference curve? We know(dy
dx
)F
= −F ′xF ′y
= −G
H
Then
y ′′ = −HG ′ − GH ′
H2
Now
G ′ =d
dx
∂F
∂x=∂2F
∂x2+
∂2F
∂y ∂x
dy
dx
=∂2F
∂x2− ∂2F
∂y ∂x
(∂F/∂x
∂F/∂y
)So
HG ′ =∂F
∂y
∂2F
∂x2− ∂2F
∂y ∂x
∂F
∂x= F ′y F ′′xx − F ′′yxF ′x
Also
H ′ =d
dx
∂F
∂y=
∂F
∂x ∂y+∂2F
∂y2
dy
dx
=∂2F
∂x ∂y− ∂2F
∂y2
(∂F/∂x
∂F/∂y
)So
GH ′ = F ′xF ′′xy −F ′′yy (F ′x)2
F ′y
=F ′xF ′′xy F ′y − F ′′yy (F ′x)2
F ′y
Putting this all together we get
y ′′ = −F ′y F ′′xx − F ′′yxF ′x −
(F ′xF ′′xy F ′y − F ′′yy (F ′x)2
F ′y
)(F ′y )2
= − 1
(F ′y )3
[F ′′xx(F ′y )2 − 2F ′′xy F ′xF ′y + F ′′yy (F ′x)2
]=
1
(F ′y )3
∣∣∣∣∣∣0 F ′x F ′y
F ′x F ′′xx F ′′xyF ′y F ′′xy F ′′yy
∣∣∣∣∣∣
Example
Along the indifference curve
1
x+
1
y= c
compute (y ′′)u. What does this say about ddx Ryx?
SolutionWe have u(x , y) = 1
x + 1y , so
(d2y
dx2
)u
=1
(−1/y2)3
∣∣∣∣∣∣0 −1/x2 −1/y2
−1/x2 2/x3 0−1/y2 0 −2/y3
∣∣∣∣∣∣
Example
Along the indifference curve
1
x+
1
y= c
compute (y ′′)u. What does this say about ddx Ryx?
SolutionWe have u(x , y) = 1
x + 1y , so
(d2y
dx2
)u
=1
(−1/y2)3
∣∣∣∣∣∣0 −1/x2 −1/y2
−1/x2 2/x3 0−1/y2 0 −2/y3
∣∣∣∣∣∣
Solution (continued)
(d2y
dx2
)u
=1
(−1/y2)3
∣∣∣∣∣∣0 −1/x2 −1/y2
−1/x2 2/x3 0−1/y2 0 −2/y3
∣∣∣∣∣∣= −y6
[−(−1
x2
)(−1
x2
)(2
y3
)−(−1
y2
)(2
x3
)(−1
y2
)]= 2y6
[1
x4y3+
1
y4x3
]= 2
(y
x
)3(
1
x+
1
y
)= 2c
(y
x
)3
This is positive, and since Ryx = −(
dydx
)u, we have
d
dxRyx = −2u
(y
x
)3< 0
So the MRS diminishes with increasing consumption of x.
Bonus: Elasticity of substitutionSee Section 16.4
The elasticity of substitution is the elasticity of the MRS withrespect to the ratio y/x:
σyx = εRyx ,(y/x) =∂Ryx
∂(y/x)·
y/x
Ryx
In our case, Ryx = (y/x)2, so
σyx = 2 (y/x)y/x
(y/x)2= 2
which is why the function u(x , y) = 1x + 1
y is called a constantelasticity of substitution function.