Lesson 19: Partial Derivatives

Transcript of Lesson 19: Partial Derivatives

Lesson 19 (Section 15.3 and 15.5)Partial Derivatives

Math 20

November 2, 2007

Announcements

I Problem Set 7 on the website. Due November 7.

I OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323)

I Prob. Sess.: Sundays 6–7 (SC B-10), Tuesdays 1–2 (SC 116)

Outline

Partial DerivativesMotivationDefinitionOther NotationsWorksheet

Second derivativesDon’t worry about the mixed partials

Marginal Quantities

Motivation

Think back to your one-variable class. What do we use thederivative for?

I Slope of the tangent line at a point

I Instantaneous rate of change at a point

I Best linear approximation near a point . . .

What is the analogue of tangent line for a function of two (ormore) variables? It’s a plane in R3 (or hyperplane in Rn+1). Whatis its “slope”?Clearly, no single scalar can describe the slope. Just by looking atthe traces which make the graph, however, we can see some“special” curves whose slopes might be significant.

Motivation

Motivation

Motivation

Motivation

Motivation

What is the analogue of tangent line for a function of two (ormore) variables?

It’s a plane in R3 (or hyperplane in Rn+1). Whatis its “slope”?Clearly, no single scalar can describe the slope. Just by looking atthe traces which make the graph, however, we can see some“special” curves whose slopes might be significant.

Motivation

What is the analogue of tangent line for a function of two (ormore) variables? It’s a plane in R3 (or hyperplane in Rn+1).

Whatis its “slope”?Clearly, no single scalar can describe the slope. Just by looking atthe traces which make the graph, however, we can see some“special” curves whose slopes might be significant.

Motivation

What is the analogue of tangent line for a function of two (ormore) variables? It’s a plane in R3 (or hyperplane in Rn+1). Whatis its “slope”?

Clearly, no single scalar can describe the slope. Just by looking atthe traces which make the graph, however, we can see some“special” curves whose slopes might be significant.

Motivation

ExampleLet f = f (x , y) = 4− x2 − 2y4. Look at the point P = (1, 1, 1)on the graph of f .

-2

-1

0

1

2

-2

-1

0

1

2

-30

-20

-10

0

-2

-1

0

1

2

z

xy

There are two interesting curves going through the point P:

x 7→ (x , 1, f (x , 1)) y 7→ (1, y , f (1, y))

Each of these is a one-variable function, so makes a curve, and hasa slope!

d

dxf (x , 1)

∣∣∣∣x=1

=d

dx

(2− x2

)∣∣∣∣x−1

= −2x |x=1 = −2.

d

dyf (1, y)

∣∣∣∣y=1

=d

dx

(3− 2y4

)∣∣∣∣y=1

= −8y3∣∣y=1

= −8.

We see that the tangent plane is spanned by these twovectors/slopes.

There are two interesting curves going through the point P:

x 7→ (x , 1, f (x , 1)) y 7→ (1, y , f (1, y))

Each of these is a one-variable function, so makes a curve, and hasa slope!

d

dxf (x , 1)

∣∣∣∣x=1

=d

dx

(2− x2

)∣∣∣∣x−1

= −2x |x=1 = −2.

d

dyf (1, y)

∣∣∣∣y=1

=d

dx

(3− 2y4

)∣∣∣∣y=1

= −8y3∣∣y=1

= −8.

We see that the tangent plane is spanned by these twovectors/slopes.

Math 20 - November 02, 2007.GWBFriday, Nov 2, 2007

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Upshot

At a point on the graph of a function of several variables, there ismore than one “slope” because there is more than one “curve”through the point. We can take a curve in each direction by(temporarily) fixing all the variables except one and treating thatlike a one-variable function. The derivative of that function is justone of the partial derivatives of f .

DefinitionLet f : Rn → R. We define the partial derivatives ∂f

∂x1, ∂f

∂x2, . . .

∂f∂xn

at a point (a1, a2, . . . , an) as

∂f

∂x1(a1, a2, . . . , an) = lim

h→0

f (a1 + h, a2, . . . , an)− f (a1, a2, . . . , an)

h,

∂f

∂x2(a1, a2, . . . , an) = lim

h→0

f (a1, a2 + h, . . . , an)− f (a1, a2, . . . , an)

h,

. . .

∂f

∂xn(a1, a2, . . . , an) = lim

h→0

f (a1, a2, . . . , an + h)− f (a1, a2, . . . , an)

h

Example

Let f (x , y) = x3 − 3xy2. Find its partial derivatives.

SolutionWhen finding ∂f

∂x , we hold y constant. So

∂f

∂x= 3x2 − (3y2)

∂

∂x(x) = 3x2 − 3y2

Similarly,∂f

∂y= 0− 3x(2y) = −6xy

Example

∂f

∂x= 3x2 − (3y2)

∂

∂x(x) = 3x2 − 3y2

Similarly,∂f

∂y= 0− 3x(2y) = −6xy

Example

∂f

∂x= 3x2 − (3y2)

∂

∂x(x) = 3x2 − 3y2

Similarly,∂f

∂y= 0− 3x(2y) = −6xy

Other Notations

∂f

∂x=

∂z

∂x=z ′x =zx =f ′x(x , y) =f ′1(x , y) =

∂f (x , y)

∂x

∂f

∂y=

∂z

∂y=z ′y =zy =f ′y (x , y) =f ′2(x , y) =

∂f (x , y)

∂y

S&H prefer the numerical subscripts rather than variable names.Other authors have different preferences.

Other Notations

∂f

∂x=

∂z

∂x=z ′x =zx =f ′x(x , y) =f ′1(x , y) =

∂f (x , y)

∂x

∂f

∂y=

∂z

∂y=z ′y =zy =f ′y (x , y) =f ′2(x , y) =

∂f (x , y)

∂y

S&H prefer the numerical subscripts rather than variable names.Other authors have different preferences.

Worksheet Problems 1–4

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Outline

Marginal Quantities

Second derivatives

If f (x , y) is a function of two variables, each of its partialderivatives are function of two variables, and we can hope thatthey are differentiable, too. So we define the second partialderivatives.

∂2f

∂x2=

∂

∂x

(∂f

∂x

)= fxx = f ′′11

∂2f

∂y ∂x=

∂

∂y

(∂f

∂x

)= fxy = f ′′12

∂2f

∂x ∂y=

∂

∂x

(∂f

∂y

)= fyx = f ′′21

∂2f

∂y2=

∂

∂y

(∂f

∂y

)= fyy = f ′′22

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Don’t worry about the mixed partials

The “mixed partials” bookkeeping may seem scary. However, weare saved by:

Theorem (Clairaut’s Theorem/Young’s Theorem)

If f is defined near (a, b) and f ′12 and f ′21 are continuous at (a, b),then

f ′′12(a, b) = f ′′21(a, b).

The upshot is that we needn’t worry about the ordering.

Don’t worry about the mixed partials

The “mixed partials” bookkeeping may seem scary. However, weare saved by:

Theorem (Clairaut’s Theorem/Young’s Theorem)

If f is defined near (a, b) and f ′12 and f ′21 are continuous at (a, b),then

f ′′12(a, b) = f ′′21(a, b).

The upshot is that we needn’t worry about the ordering.

Example (Continued)

Let f (x , y) = x3 − 3xy2. Find the second derivatives of f .

SolutionWe have

f ′′11 = (3x2 − 3y2)x = 6x

f ′′12 = (3x2 − 3y2)y = −6y

f ′′21 = (−6xy)x = −6y

f ′′22 = (−6xy)y = −6x

Notice that f ′′21 = f ′′12, as predicted by Clairaut (everything is apolynomial here so there are no concerns about continuity). Thefact that f ′′11 = −f ′′22 is a coincidence.

Example (Continued)

SolutionWe have

f ′′11 = (3x2 − 3y2)x = 6x

f ′′12 = (3x2 − 3y2)y = −6y

f ′′21 = (−6xy)x = −6y

f ′′22 = (−6xy)y = −6x

Example (Continued)

SolutionWe have

f ′′11 = (3x2 − 3y2)x = 6x

f ′′12 = (3x2 − 3y2)y = −6y

f ′′21 = (−6xy)x = −6y

f ′′22 = (−6xy)y = −6x

Worksheet Problems 5–6

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Outline

Marginal Quantities

Marginal Quantities

If a variable u depends on some quantity x , the amount that uchanges by a unit increment in x is called the marginal u of x .For instance, the demand q for a quantity is usually assumed todepend on several things, including price p, and also perhapsincome I . If we use a nonlinear function such as

q(p, I ) = p−2 + I

to model demand, then the marginal demand of price is

∂q

∂p= −2p−3

Similarly, the marginal demand of income is

∂q

∂I= 1