M38 Lec 021314

MATH 38: Mathematical Analysis IIIUnit 3: Differentiation of Functions of More than One Variable

I. F. Evidente

IMSP (UPLB)

Outline

1 Differentiability

2 Local Linear Approximation

3 The Total Differential

4 Partial Derivates as Rate of Change

Figures taken from: J. Stewart, The Calculus: Early Transcendentals,Brooks/Cole, 6th Edition, 2008.

Given: f (z) where z is a function of x and y .

So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz z

x= cos(x2+ y2) 2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz z

x= cos(x2+ y2) 2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz z

x= cos(x2+ y2) 2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz z

x= cos(x2+ y2) 2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)=

cosz zx

= cos(x2+ y2) 2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz

z

x= cos(x2+ y2) 2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz z

x=

cos(x2+ y2) 2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz z

x= cos(x2+ y2)

2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz z

x= cos(x2+ y2) 2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz z

x= cos(x2+ y2) 2x

If we get the composition first:

f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz z

x= cos(x2+ y2) 2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz z

x= cos(x2+ y2) 2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]=

cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz z

x= cos(x2+ y2) 2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2)

2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz z

x= cos(x2+ y2) 2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz z

x= cos(x2+ y2) 2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: f (z) where z is a function of x and y . So z depends on x.

x[ f (z)]= f (z)z

x

ExampleSuppose f (z)= sinz, where z = x2+ y2.

x(sinz)= cosz z

x= cos(x2+ y2) 2x

If we get the composition first: f (x, y)= sin(x2+ y2)

x[sin(x2+ y2]= cos(x2+ y2) 2x

If z is dependent on x, apply chain rule.

Given: w = f (x, y,z).

Here, z is also an independent variable! It does notdepend on x.

ExampleSuppose w = x2+ y2+ z2

w

x= 2x

If z is independent of x, then treat z like a constant when getting thepartial derivative with respect to x.

Given: w = f (x, y,z). Here, z is also an independent variable!

It does notdepend on x.

ExampleSuppose w = x2+ y2+ z2

w

x= 2x

If z is independent of x, then treat z like a constant when getting thepartial derivative with respect to x.

Given: w = f (x, y,z). Here, z is also an independent variable! It does notdepend on x.

ExampleSuppose w = x2+ y2+ z2

w

x= 2x

If z is independent of x, then treat z like a constant when getting thepartial derivative with respect to x.

Given: w = f (x, y,z). Here, z is also an independent variable! It does notdepend on x.

ExampleSuppose w = x2+ y2+ z2

w

x= 2x

If z is independent of x, then treat z like a constant when getting thepartial derivative with respect to x.

Given: w = f (x, y,z). Here, z is also an independent variable! It does notdepend on x.

ExampleSuppose w = x2+ y2+ z2

w

x=

2x

If z is independent of x, then treat z like a constant when getting thepartial derivative with respect to x.

Given: w = f (x, y,z). Here, z is also an independent variable! It does notdepend on x.

ExampleSuppose w = x2+ y2+ z2

w

x= 2x

If z is independent of x, then treat z like a constant when getting thepartial derivative with respect to x.

Given: w = f (x, y,z). Here, z is also an independent variable! It does notdepend on x.

ExampleSuppose w = x2+ y2+ z2

w

x= 2x

If z is independent of x, then treat z like a constant when getting thepartial derivative with respect to x.

Outline

1 Differentiability

2 Local Linear Approximation

3 The Total Differential

4 Partial Derivates as Rate of Change

DefinitionA function of 2 variables f (x, y) is said to be differentiable at (x0, y0) ifthe following conditions are satisfied:

1 fx(x0, y0) and fy (x0, y0) exist, and

2 lim(x,y)(x0,y0)

f (x, y)L(x, y)(xx0)2+ (y y0)2

= 0, where

L(x, y)= f (x0, y0)+ fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)

If f is differentiable at (x0, y0), L(x, y) is called the local linearapproximation of f at (x0, y0)

DefinitionA function f (x, y) is said to be differentiable if f (x, y) is differentiable forall (x, y) R2.

TheoremIf a function is differentiable at a point, then it is continuous at that point.

DefinitionA function f (x, y) is said to be differentiable if f (x, y) is differentiable forall (x, y) R2.

TheoremIf a function is differentiable at a point, then it is continuous at that point.

RemarkProperties of a function f that is differentiable at (x0, y0):

1 The graph of f has a non-vertical tangent plane at (x0, y0, f (x0, y0)).An equation of this tangent plane is z = L(x, y).

2 L(x, y) approximates f (x, y) when (x, y) is very close to (x0, y0).

f (x, y) L(x, y)

3 f is continuous at (x0, y0).

RemarkProperties of a function f that is differentiable at (x0, y0):

1 The graph of f has a non-vertical tangent plane at (x0, y0, f (x0, y0)).An equation of this tangent plane is z = L(x, y).

2 L(x, y) approximates f (x, y) when (x, y) is very close to (x0, y0).

f (x, y) L(x, y)

3 f is continuous at (x0, y0).

RemarkProperties of a function f that is differentiable at (x0, y0):

1 The graph of f has a non-vertical tangent plane at (x0, y0, f (x0, y0)).An equation of this tangent plane is z = L(x, y).

2 L(x, y) approximates f (x, y) when (x, y) is very close to (x0, y0).

f (x, y) L(x, y)

3 f is continuous at (x0, y0).

RemarkProperties of a function f that is differentiable at (x0, y0):

1 The graph of f has a non-vertical tangent plane at (x0, y0, f (x0, y0)).An equation of this tangent plane is z = L(x, y).

2 L(x, y) approximates f (x, y) when (x, y) is very close to (x0, y0).

f (x, y) L(x, y)

3 f is continuous at (x0, y0).

RemarkProperties of a function f that is differentiable at (x0, y0):

1 The graph of f has a non-vertical tangent plane at (x0, y0, f (x0, y0)).An equation of this tangent plane is z = L(x, y).

2 L(x, y) approximates f (x, y) when (x, y) is very close to (x0, y0).

f (x, y) L(x, y)

3 f is continuous at (x0, y0).

TheoremIf all first-order partial derivatives of f exist and are continuous at a point,then f is differentiable at that point.

ExampleBy the theorem, f (x, y)= x2+ y2 is differentiable.

RemarkAlso by the theorem:

1 A polynomial functions are differentiable.2 A rational functions is differentiable at every point in its domain.

TheoremIf all first-order partial derivatives of f exist and are continuous at a point,then f is differentiable at that point.

ExampleBy the theorem, f (x, y)= x2+ y2 is differentiable.

RemarkAlso by the theorem:

1 A polynomial functions are differentiable.2 A rational functions is differentiable at every point in its domain.

TheoremIf all first-order partial derivatives of f exist and are continuous at a point,then f is differentiable at that point.

ExampleBy the theorem, f (x, y)= x2+ y2 is differentiable.

RemarkAlso by the theorem:

1 A polynomial functions are differentiable.2 A rational functions is differentiable at every point in its domain.

Outline

1 Differentiability

2 Local Linear Approximation

3 The Total Differential

4 Partial Derivates as Rate of Change

DefinitionIf f is differentiable at (x0, y0), then the local linear approximation of fat (x0, y0) is

L(x, y)= fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)

RemarkThe graph of L(x, y) is the tangent plane to the surface z = f (x, y) at thepoint (x0, y0).

RemarkWhen (x, y) is very close to (x0, y0), then f (x, y) L(x, y).

DefinitionIf f is differentiable at (x0, y0), then the local linear approximation of fat (x0, y0) is

L(x, y)= fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)

RemarkThe graph of L(x, y) is the tangent plane to the surface z = f (x, y) at thepoint (x0, y0).

RemarkWhen (x, y) is very close to (x0, y0), then f (x, y) L(x, y).

DefinitionIf f is differentiable at (x0, y0), then the local linear approximation of fat (x0, y0) is

L(x, y)= fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)

RemarkThe graph of L(x, y) is the tangent plane to the surface z = f (x, y) at thepoint (x0, y0).

RemarkWhen (x, y) is very close to (x0, y0), then f (x, y) L(x, y).

Example

The function f (x, y)= 1x2+ y2

is differentiable at (4,3).

1 Find the local linear approximation of f (x, y)= 1x2+ y2

at (4,3).

2 Approximate f (4.01,2.99) using L(x, y).

Answer:L(4.01,2.00)= 0.202 and f (4.01,2.99)= 0.200

Example

The function f (x, y)= 1x2+ y2

is differentiable at (4,3).

1 Find the local linear approximation of f (x, y)= 1x2+ y2

at (4,3).

2 Approximate f (4.01,2.99) using L(x, y).

Answer:L(4.01,2.00)= 0.202 and f (4.01,2.99)= 0.200

Outline

1 Differentiability

2 Local Linear Approximation

3 The Total Differential

4 Partial Derivates as Rate of Change

Let z = f (x, y) such that f is differentiable at (x0, y0).

Suppose (x, y) changes from (x0, y0) to (x1, y1).Note: Change is always (new)-(original)

Changes in independent variables:Change in x: x = x1x0Change in y : y = y1 y0

What is the effect of this change on the dependent variable z?

Change in the dependent variable:

z = f (x1, y1) f (x0, y0)

Let z = f (x, y) such that f is differentiable at (x0, y0).Suppose (x, y) changes from (x0, y0) to (x1, y1).

Note: Change is always (new)-(original)

Changes in independent variables:Change in x: x = x1x0Change in y : y = y1 y0

What is the effect of this change on the dependent variable z?

Change in the dependent variable:

z = f (x1, y1) f (x0, y0)

Let z = f (x, y) such that f is differentiable at (x0, y0).Suppose (x, y) changes from (x0, y0) to (x1, y1).Note: Change is always (new)-(original)

Changes in independent variables:Change in x: x = x1x0Change in y : y = y1 y0

What is the effect of this change on the dependent variable z?

Change in the dependent variable:

z = f (x1, y1) f (x0, y0)

Let z = f (x, y) such that f is differentiable at (x0, y0).Suppose (x, y) changes from (x0, y0) to (x1, y1).Note: Change is always (new)-(original)

Changes in independent variables:Change in x: x = x1x0Change in y : y = y1 y0

What is the effect of this change on the dependent variable z?

Change in the dependent variable:

z = f (x1, y1) f (x0, y0)

Let z = f (x, y) such that f is differentiable at (x0, y0).Suppose (x, y) changes from (x0, y0) to (x1, y1).Note: Change is always (new)-(original)

Changes in independent variables:Change in x: x = x1x0Change in y : y = y1 y0

What is the effect of this change on the dependent variable z?

Change in the dependent variable:

z = f (x1, y1) f (x0, y0)

Let z = f (x, y) such that f is differentiable at (x0, y0).Suppose (x, y) changes from (x0, y0) to (x1, y1).Note: Change is always (new)-(original)

Changes in independent variables:Change in x: x = x1x0Change in y : y = y1 y0

What is the effect of this change on the dependent variable z?

Change in the dependent variable:

z = f (x1, y1) f (x0, y0)

As in the function of one-variable case:

Differentials of the independent variables

dx =x and dy =y

DefinitionIf z = f (x, y) is differentiable at (x0, y0), then the total differential of z at(x0, y0) is given by

dz = fx(x0, y0)dx+ fy (xo , y0)dy

RemarkGeneral formula for the differential at (x, y)

dz = fx(x, y)dx+ fy (x, y)dy

DefinitionIf z = f (x, y) is differentiable at (x0, y0), then the total differential of z at(x0, y0) is given by

dz = fx(x0, y0)dx+ fy (xo , y0)dy

RemarkGeneral formula for the differential at (x, y)

dz = fx(x, y)dx+ fy (x, y)dy

ExampleCompute dz for the following:

1 z = x2+ y22 z = tan1(xy)

Extend to functions of more than two variables.

ExampleLet w = x2y4z3+xy + z2+1. Find dw .

ExampleCompute dz for the following:

1 z = x2+ y22 z = tan1(xy)

Extend to functions of more than two variables.

ExampleLet w = x2y4z3+xy + z2+1. Find dw .

Consider the expressionsdz

dxand

z

x

RemarkNote that dz and dx by themselves make sense, but z and x bythemselves make no sense.

Consider the expressionsdz

dxand

z

x

RemarkNote that dz and dx by themselves make sense, but z and x bythemselves make no sense.

Let L(x, y) be the local linear approximation of f (x, y) at (x0, y0)

f (x, y) L(x, y)f (x, y) fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)

f (x, y) f (x0, y0) fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)z dz

Note: In the formula dz = fx(x, y)dx+ fy (x, y)dy , x and y are the originalvalues of x and y .

Let L(x, y) be the local linear approximation of f (x, y) at (x0, y0)

f (x, y) L(x, y)

f (x, y) fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)f (x, y) f (x0, y0) fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)

z dz

Note: In the formula dz = fx(x, y)dx+ fy (x, y)dy , x and y are the originalvalues of x and y .

Let L(x, y) be the local linear approximation of f (x, y) at (x0, y0)

f (x, y) L(x, y)f (x, y) fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)

f (x, y) f (x0, y0) fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)z dz

Note: In the formula dz = fx(x, y)dx+ fy (x, y)dy , x and y are the originalvalues of x and y .

Let L(x, y) be the local linear approximation of f (x, y) at (x0, y0)

f (x, y) L(x, y)f (x, y) fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)

f (x, y) f (x0, y0) fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)

z dz

Note: In the formula dz = fx(x, y)dx+ fy (x, y)dy , x and y are the originalvalues of x and y .

Let L(x, y) be the local linear approximation of f (x, y) at (x0, y0)

f (x, y) L(x, y)f (x, y) fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)

f (x, y) f (x0, y0) fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)z dz

Note: In the formula dz = fx(x, y)dx+ fy (x, y)dy , x and y are the originalvalues of x and y .

Let L(x, y) be the local linear approximation of f (x, y) at (x0, y0)

f (x, y) L(x, y)f (x, y) fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)+ f (x0, y0)

f (x, y) f (x0, y0) fx(x0, y0)(xx0)+ fy (x0, y0)(y y0)z dz

Note: In the formula dz = fx(x, y)dx+ fy (x, y)dy , x and y are the originalvalues of x and y .

ExampleUse differentials to approximate the change in f (x, y)= x2+ y2 as (x, y)changes from (4,3) to (4.01,2.99).

Answer:dz = 0.02 and z = 0.0202

ExampleUse differentials to approximate the change in f (x, y)= x2+ y2 as (x, y)changes from (4,3) to (4.01,2.99).

Answer:dz = 0.02 and z = 0.0202

ExampleA metal container in the shape of a right circular cylinder with a height of6 inches and a radius of 2 inches is to be coated in material 0.1 inchesthick. Approximate the amount of coating material to be used.

Example

A open wooden rectangular box is to be made of lumber that is 23 inchthick. The inside length is to be 6 inches, the inside width is to be 3 inchesand the inside depth is to be 4 inches. Approximate the amount of lumberto be used in the box.

(amount means volume of material)

Example

A open wooden rectangular box is to be made of lumber that is 23 inchthick. The inside length is to be 6 inches, the inside width is to be 3 inchesand the inside depth is to be 4 inches. Approximate the amount of lumberto be used in the box. (amount means volume of material)

Outline

1 Differentiability

2 Local Linear Approximation

3 The Total Differential

4 Partial Derivates as Rate of Change

Physical Interpretation of Partial DerivativesSuppose z is a function of x and y .

z

xis the instantaneous rate of change in z per unit change in x as y

is held constant.z

yis the instantaneous rate of change in z per unit change in y as x

is held constant.

ExampleSuppose that the weekly profit of a store is given byP (x, y)= 3000+240y 20y(x2y)10(x2)2, where x = V10,000 , V is thevalue of the inventory in the store, and y is the number of clerks employed.At present the value of the inventory in the story is 180,000 pesos andthere are 8 clerks. Find the rate at which P is changing with respect to xat this instant if y remains fixed at 8.

Interpretation of answer:At this instant, if x is increased by 1 (or V is increased by 10,000) andretain 8 clerks, then the profit will increase by 160 pesos.

ExampleSuppose that the weekly profit of a store is given byP (x, y)= 3000+240y 20y(x2y)10(x2)2, where x = V10,000 , V is thevalue of the inventory in the store, and y is the number of clerks employed.At present the value of the inventory in the story is 180,000 pesos andthere are 8 clerks. Find the rate at which P is changing with respect to xat this instant if y remains fixed at 8.

Interpretation of answer:At this instant, if x is increased by 1 (or V is increased by 10,000) andretain 8 clerks, then the profit will increase by 160 pesos.

ExampleThe temperature at any point (x, y) on a flat plate isT (x, y)= 54 23x24y2 degrees Celsius. Find the rate at which thetemperature is changing in the direction of the positive y-axis at the point(3,1).

Interpretation of answer:If one travels in the direction of the positive y axis from (3,1), thetemperature decreases by 8 degrees Celsius per unit change in y .

ExampleThe temperature at any point (x, y) on a flat plate isT (x, y)= 54 23x24y2 degrees Celsius. Find the rate at which thetemperature is changing in the direction of the positive y-axis at the point(3,1).

Interpretation of answer:If one travels in the direction of the positive y axis from (3,1), thetemperature decreases by 8 degrees Celsius per unit change in y .

AnnouncementsRecall Class Policy:

1 Missed midterm, prefinal or final exam, with valid excuse: SpecialExam (Harder than the usual)

2 Missed Chapter Quiz, with valid excuse: 1st missed chapter quiz willnot be included in total, succeeding missed chapter quizzes grade of 0

DifferentiabilityLocal Linear ApproximationThe Total DifferentialPartial Derivates as Rate of Change