lesson 24: implicit differentiation

Lesson 24 (Sections 16.3, 16.7)Implicit Differentiation

Math 20

November 16, 2007

Announcements

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I There will be class November 21 and homework dueNovember 28.

I next OH: Monday 1-2pm, Tuesday 3-4pm

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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.