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SIMPLE MULTIVARIATE CALCULUS

1. REAL-VALUED FUNCTIONS OF SEVERAL VARIABLES

1.1. Definition of a real-valued function of several variables. Suppose D is a set of n-tuples of real num-bers (x1, x2, x3, . . . , xn). A real-valued function f on D is a rule that assigns a single real number

y = f (x1, x2, x3, . . . , xn)to each element in D. The set D is the function’s domain. The set of y-values taken on by f is the range of

the function. The symbol y is the dependent variable of f, and f is said to be a function of the n independentvariables x1 to xn. We also call the x’s the function’s input variables and we call y the function’s outputvariable.

A real-valued function of two variables is just a function whose domain is R2 and whose range is asubset of R1, or the real numbers. If we view the domain D as column vectors in Rn, we sometimes writethe function as

f

x1

x2

...xn

1.2. Examples.a: The volume of a right circular cylinder is a function of the radius and height, V = f(r, h) or V =

πr2h. A cylinder is represented in figure 1. The volume of the cylinder as a function of its radiusand height is shown graphically in figure 2. Notice that the volume increases as both the radiusand the height increase.

FIGURE 1. A Cylinder

Date: March 22, 2006.1

2 SIMPLE MULTIVARIATE CALCULUS

FIGURE 2. The Volume of a Cylinder as a Function of Radius and Height

0

1

2

3r

0

1

2

3h

0

20

40

60

80

VHr,hL

0

1

2r

0

1

2h

b: The level of production from a given technology as a function of two inputs x1 and x2 is repre-sented by y = f(x1, x2). For example, a quadratic function might be f(x1, x2) = 20x1 + 16x2 − 2x2

1 − x22.

A Cobb-Douglas function might be f(x1, x2) = 10x1/41 x

1/22 . We can represent the quadratic func-

tion with a graph in three dimensions as in figure 3. We can represent the Cobb-Douglas functionwith a graph in three dimensions as in figure 4.

FIGURE 3. Quadratic Production Function f(x1, x2) = 20x1 + 16x2 − 2x21 − x2

2

0

2

4

6

8

x1

0

5

10

x2

025

50

75

100

fHx1,x2L

2

4

6

8

x1

1.3. Interior and boundary points in the plane or R2.

1.3.1. Interior points of regions in the plane (R2). A point (x01, x0

2) in a region B in the x1-x2 plane is an interiorpoint of B if it is the center of a disk that lies entirely in B.

1.3.2. Boundary points of regions in the plane (R2). A point (x0, y0) is a boundary point of B if every diskcentered at (x0, y0) contains points that lie outside of B and well as points that lie in B. (The boundary pointitself needs to belong to B).

The interior points of a region, as a set, make up the interior of the region. The region’s boundary pointsmake up its boundary.

SIMPLE MULTIVARIATE CALCULUS 3

FIGURE 4. Cobb-Douglas Production Function f(x1, x2) = 10x1/41 x

1/22

10

20

30

x1

10

20

30

x2

0

50

100f Hx1,x2L

10

20x1

1.3.3. Open and closed regions in the plane (R2). A region is open if it consists entirely of interior points. Aregion is closed if it contains all of its boundary points.

1.3.4. Bounded regions in the plane. A region in the plane is bounded if it lies inside a disk of fixed radius. Aregion is unbounded if it is not bounded.

Consider the region in the plane bounded by line segments in igure 5. Call this region the set B. The pointa is an interior point because all points in a small disk centered at a lie within B. The point c is a boundarypoint because no matter how small the radius of the disk centered at c, it contains points both within andoutside of B. The boundary of B consists of all points on the line segments defining B, and the interior of Bconsists of all points bounded by the line segments but which do not lie on them. The set B is a closed set.

Consider the region in the plane bounded by the dotted line in figure 6. Call this region B. The set consistsall points within the dotted circle but does not include points on the circle. The point a is an interior pointbecause all points in a small disk centered at a lie within B. The point c is a boundary point because nomatter how small the radius of the disk centered at c, it contains points both within and outside of B. In thiscase the boundary point c is not contained in the set B. The boundary of B consists of all points on the circledefining B, and the interior of B consists of all points bounded by circle which do not lie on it. The set B isan open set.

Consider the region in the plane on and to the northeast of the curved line in figure 7. Call this regionB. The point a is an interior point because all points in a small disk centered at a lie within B. The point c isa boundary point because no matter how small the radius of the disk centered at c, it contains points bothwithin and outside of B. The interior of B consists of all points to the northeast of the curved line in figure7. The set B is an unbounded set. The set B is also a closed set.

1.4. Interior and boundary points in space or R3.

1.4.1. Interior points of regions in space (R3). A point (x01, x0

2, x03) in a region D in space is n interior point of

D if it is the center of a ball that lies entirely in D.


FIGURE 5. Interior and Boundary Points of a Region in the Plane

x1

x2

B

0

ac


x1

x2

B

0

a

c

1.4.2. Boundary points of regions in space (R3). A point (x01, x0

2, x03) is a boundary point of D f every sphere

centered at (x01, x0

2, x03) encloses points that lie outside of D and well as points that lie in D.



x1

x2

0

ac

B

The interior of D is the set of interior point of D. The boundary of D is the set of boundary points of D.

1.4.3. Open and closed regions in space(R3). A region D is open if it consists entirely of interior points. Aregion is closed if it contains its entire boundary.

1.5. Interior and boundary points in Rn. We can extend the idea of open and closed sets to Rn. Let a =(a1,,a2,. . . ,an) be a point in Rn and let r be a given positive number. The set of all points x ∈ Rn such that

( [x− a] · [x− a] )1/2< r

is called an open n-ball of radius r with center at a. We call ( [x− a] · [x− a] )1/2 the distance betweenthe point x and the point a. We denote this set by B(a;r). The ball B(a;r) consists of all points whose distancefrom a is less than r. In R1, this is an open interval with center at a. In R2, this is a circular disk with radiusr and center at a. In R3, this is spherical solid with center at a and radius r.

1.5.1. Interior points of sets in Rn. Let S be a subset of Rn, and assume that the point a is an element of S.Then a is called an interior point of S if there is an open n-ball with center at a, all of whose points belongto S.

1.5.2. Open sets in Rn. A set S in Rn is called open if all its points are interior points.

1.5.3. Closed sets in Rn. A set S in Rn is called closed if its complement Rn \ S is open.

Consider the region in space consisting of the doughnut or torus in figure 8. Call this region B in figure9. The point a is an interior point because all points in a ball or sphere centered at a lie within B as shownin figure 10. The point c is not an interior point because a sphere centered at point c would contain pointsboth inside and outside the doughnut.

1.6. Graphs and level curves for functions with domain in R2.


FIGURE 8. A Subset of R3

FIGURE 9. The Interior of a Subset of R3

B

a

c

1.6.1. The graph of a function with domain in R2. The set of all points (x1, x2, f(x1, x2)) in space (R3), for (x1,x2) in the domain of f, is called the graph of f. The graph of f is also called the surface z = f(x1, x2). Considerthe function

z = f (x1, x2) = 100 − x21 − x2

2

This function has a graph in R3 because its domain is R2. The graph of f(x1, x2) = 100 − x21 − x2

2 isshown in figure 11.

1.6.2. The level curve of a function with domain in R2. The set of points in the plane (R2) where a function f(x1,x2) has a constant value f(x1, x2) = c is called a level curve of f. One can think of a level curve like contourlines on a map. Points on the same curve or line represent the same height of the function. Level curves arecreated by intersecting a plane at a given height (or value of the function f(x1,x2)) with the graph of f(x1,x2),noting the values of (x1,x2) where these intersections occur, and then plotting them in the x1-x2 plane. Asan example, consider the function

z = f (x1, x2) = 100 − x21 − x2

2

from figure 11. In figure 12, we show both the graph of the function and a plane at function value of 30.

A series of level curves for f(x1, x2) = 100 − x21 − x2

2 at various valued of the function are contained infigure 13.


FIGURE 10. An Interior Point of a Subset of R3 is Contained within a Sphere within the Subset

B

a

c

FIGURE 11. Graph of the function z = 100 − x21 − x2

2

-5

0

5

10

x1

-10

-5

0

5

10

x2

-100-50

0

50

100

f Hx1,x2L

-5

0

5x1

1.6.3. Second example of a function with domain in R2. Consider the function

z = f (x1, x2) =x2

1 x22

(1 + x21) (1 + x2

2)


FIGURE 12. Graph of the function f(x1, x2) = 100 − x21 − x2

2 along with intersecting plane

-10

-5

0

5

10

x1

-10

-5

0

5

10

x2

0

50

100

f Hx1,x2L

-10

-5

0

5

10

x1

FIGURE 13. Contour lines for the function f(x1, x2) = 100 − x21 − x2

2

-10 -5 0 5 10-10

-5

0

5

10

The graph of the function and a plane at z = 2 is depicted in figure 14. A level curve for the functionwhen z = 2 is given in figure 15. A more general set of level curves is depicted in figure 16.

1.6.4. Third example of a function with domain in R2. Consider the function


FIGURE 14. Graph of the function z = x21 x2

2(1 + x2

1) (1 + x22)

00.5

1.5

2

x1

0 0.5 1 1.5 2

x2

0

1

2

3

f Hx1,x2L

0

1

FIGURE 15. Contour line for the function z = x21 x2

2(1 + x2

1) (1 + x22)

when z = 2

0 0.5 1 1.5 2x1

0

0.5

1

1.5

2

x2

z = f (x1, x2) =−x1 x2

ex21 + x2

2

The graph of the function is depicted in figure 17. A set of level curves is depicted in figure 18.

1.6.5. Functions with domain in R3. The set of points (x1, x2, x3) in space where a function of three indepen-dent variables f(x1, x2, x3) has a constant value f(x1, x2, x3) = c is called a level surface of f. The set of allpoints [x1, x2, x3, f(x1, x2, x3)] in space, for (x1, x2, x3) in the domain of f, is called the graph of f. The graphof f is also called the surface w = f(x1, x2, x3). One example of the level set for a function with a domain inR3 in contained in figure 19 while a second example is shown in figure 20.

2. LIMITS

2.1. Definition of a limit for a real valued function with a domain in R2.

Definition 1 (Limit). We say that a function f(x1,x2) approaches the limit L as (x1, x2) approaches (x01, x0

2),and write


FIGURE 16. Contour lines for the function z = x21 x2

2(1 + x2

1) (1 + x22)

0 0.5 1 1.5 2x1

0

0.5

1

1.5

2

x2

FIGURE 17. Graph of the function f(x1, x2) = −x1 x2

ex21 + x2

2

-2

-1

0

1

2

x1

-2

-1

0

1

2

x2

-0.2

-0.1

0

0.1

0.2

fHx1,x2L-2

-1

0

1

2

x1

-

-

0

lim(x1, x2 ) → (x0

1, x02 )

f(x1, x2) = L

if, for every number ε > 0, there exists a corresponding number δ > 0 such that for all (x1,x2) in thedomain of f,

0 <√

(x1 − x01 )2 + (x2 − x0

2)2 < δ ⇒ | f(x1, x2) − L | < ε

Notice that | f(x1, x2 )−L | is the distance between the numbers f(x1, x2) and L, and√

(x1 − x01 )2 + (x2 − x0

2)2is the distance between the point (x1,x2) and the point (x0

1,x02). Thus definition 1 says that the distance be-

tween f(x1, x2) and L can be made arbitrarily small by making the distance between (x1,x2) and (x01,x0

2)sufficiently small (but not zero).


FIGURE 18. Levels curves for the function f(x1, x2) = −x1 x2

ex21 + x2

2

-2 -1 0 1 2x1

-2

-1

0

1

2x2

FIGURE 19. Contour lines for the function f(x1, x2, x3) = x21 + 2x2

2 + x2x3

-1

0

1x1

-1

-0.5

0

0.5

1

x2

-1

-0.5

0

0.5

1

x3

-1

0

1x1

Figure 21 shows a rectangular neighborhood for a point in the x1-x2 plane and the corresponding portionof the surface. The minimum value that the function achieves over the plane in denoted ”min” on thevertical axis, while the maximum value that the function achieves over the plane in denoted ”max” on thevertical axis. The idea of a limit is that as we make the rectangle smaller and smaller, the difference betweenthe ”min” and ”max” points will get smaller and smaller. Formally, we consider the distance of points froma given point in the domain of the function (a disc in the case of R2), as the diameter of this disk approacheszero, the value of the function will approach a fixed number.

The definition of limit applies to boundary points (x01, x0

2) as well as interior points of the domain of f.The only requirement is that the point (x1, x2) remain in the domain at all times. It can be shown that


FIGURE 20. Contour lines for the function f(x1, x2, x3) = 10x1/31 x

2/52 x

1/43

0 24

x1

0 2 4

x2

0

2

4

x3

0

2

4

FIGURE 21. Finding the Limiting Value for the Function f(x1, x2) = 10x1/41 x

1/22 as (x1,x2)→ (17,17)

10

20

30

x1

010

2030x2

25

50

75

125

fmin

fmax

fHx1,x2L

25

50

L

lim(x1, x2)→(x0

1, x02)

x1 = x01 (1a)

lim(x1, x2)→(x0

1, x02)

x2 = x02 (1b)

lim(x1, x2)→(x0

1, x02)

k = k (1c)


For example in equation 1a, f(x1,x2) = x1 and L = x01. Suppose ε > 0 is chosen and let δ = ε. Then using

the definition of limit we see that

0 <√

(x1 − x01 )2 + (x2 − x0

2)2 < δ = ε

⇒√

(x1 − x01 )2 < ε

⇒ |(x1 − x01 )| < ε,

√a2 = |a|

⇒ |(f(x1, x2) − x01 )| < ε, x1 = f(x1, x2)

That is

|(f(x1, x2) − x01 )| < ε whenever 0 <

√(x1 − x0

1 )2 + (x2 − x02)2 < δ

So

lim(x1, x2)→(x0

1, x02)

f(x1, x2) = lim(x1, x2)→(x0

1, x02)

x1 = x01

2.2. Properties of limits.

Theorem 1 (Properties of Limits of Functions with Domain in R2).

The following rules hold if L, M, and k are real numbers and

lim(x1, x2 ) → (x0

1, x02 )

f(x1, x2) = L

lim(x1, x2 ) → (x0

1, x02 )

g(x1, x2) = M

lim(x1, x2 ) → (x0

1, x02 )

[f(x1, x2) + g(x1, x2)] = L + M (2a)

lim(x1, x2 ) → (x0

1, x02 )

[f(x1, x2)− g(x1, x2)] = L−M (2b)

lim(x1, x2 ) → (x0

1, x02 )

[f(x1, x2)g(x1, x2)] = L ·M (2c)

lim(x1, x2 ) → (x0

1, x02 )

[k f(x1, x2)] = k L (2d)

lim(x1, x2 ) → (x0

1, x02 )

[f(x1, x2)g(x1, x2)

]=

L

M, M 6= 0 (2e)

If r and s are integers with no common factors, and s 6= 0 then

lim(x1, x2 ) → (x0

1, x02 )

[f(x1, x2)]rs = L

rs (provided L

rs is defined) (2f)

If s is an even number, then it is assumed that L > 0


3. CONTINUITY

A function f(x1,x2) is continuous at the point (x01, x0

2) if1. f is defined at (x0

1, x02)

2. lim(x1, x2 ) → (x01, x0

2)f(x1, x2) exists

3. lim(x1, x2 ) → (x01, x0

2)f(x1, x2) = f(x0

1, x02)

The intuitive meaning of continuity is that if the point (x1,x2) changes by a small amount, then the valueof f(x1,x2) changes by a small amount. The means that the surface which is the graph of a continuousfunction has no hole or break.

Sums, differences, products and quotients of continuous functions are continuous on their domains.

4. DEFINITION OF PARTIAL DIFFERENTIATION

Let f be a function with domain an open set in Rn and range in R1, i.e. y = f(x1, x2, ... , xn). Now definethe partial derivative of f with respect to xi as

∂f(x)∂xi

= fi(x) = limh→ 0

f(x1, x2, ...xi + h, ..., xn) − f(x1, x2, ...xi, ..., xn)h

(3)

whenever the limit exists.

The slope of the curve z = f(x1, x2) at the point (x01, x0

2, f(x01, x0

2)) in the plane x2 = x02 is the value of the

partial derivative of f with respect to x at (x01, x0

2).

5. USING THE LIMIT CONCEPT TO COMPUTE A PARTIAL DERIVATIVE

5.1. Procedure.a: Add the vector h with zeros in all but one place to the vector x (h6=0) and compute f(x+h).b: Compute f(x)c: Compute the change in the value function: f(x+h) - f(x).d: For h 6= 0, form the quotient f (x + h) − f (x)

h .e: Simplify the fraction in d as much as possible. Whenever possible, cancel h from both the numer-

ator and denominator.f: f´(x) is the number that f (x + h) − f (x)

h approaches as h tends to zero.

5.2. Example. Let the function bef(x1, x2) = 2x3

1 (x22 + 1)

and compute the partial derivative of f(x1, x2) with respect to x1.a: f (x + h ) = 2 (x1 + h)3 (x2

2 + 1) = 2 (x31 + 3 x2

1 h + 3 h2 x1 + h3) (x22 + 1)

b: f (x) = 2 x31 (x2

2 + 1)

c: f (x + h ) − f (x) = 2 ( 3 x21 h + 3 h2 x1 + h3) (x2

2 + 1)

d:f (x + h) − f (x)

h= 2 ( 3 x2

1 h + 3 h2 x1 + h3) (x22 + 1)

h

e:2 ( 3 x2

1 h + 3 h2 x1 + h3) (x22 + 1)

h= 2 ( 3 x2

1 + 3 h x1 + h2) (x22 + 1)

f: As h → 0, the expression goes to 6 x21 (x2

2 + 1) .


6. CALCULATING PARTIAL DERIVATIVES

6.1. The intuitive idea of computing a partial derivative. We calculate ∂ f∂ x1

by differentiating f with respectto x1 in the usual way while treating x2 as a constant. Similarly for other partial derivatives.

6.2. Example problems.a.

f(x1, x2) = x21 x2 + 8x2

1 x32 + x1 ln(x2)

∂f(x1, x2)∂x1

= 2x1 x2 + 16x1 x32 + ln(x2)

∂f(x1, x2)∂x2

= x21 + 24x2

1 x22 +

x1

x2

b.f (x1, x2, x3) = x1 sin (x2 + 3x3)

∂f

∂x1= sin (x2 + 3x3)

∂ f

∂x2= x1 cos (x2 + 3x3)

∂ f

∂ x3= 3x1 cos(x2 + 3x3)

c.f (x1, x2, x3) = 50 + 5x1 + 3 x2 + 2 x3 + 7x2

1

+ 2 x1x2 + 3x1 x3 + 5 x22 + 4 x2 x3 + 2x2

3

∂f

∂x1= 5 + 14 x1 + 2 x2 + 3 x3

∂ f

∂ x2= 3 + 2 x1 + 10 x2 + 4 x3

∂ f

∂ x3= 2 + 3 x1 + 4 x2 + 4 x3

7. GEOMETRIC INTERPRETATION OF PARTIAL DERIVATIVES

When the domain of a function is R2 and the graph of the function is in R3, a partial derivative withrespect to one of the variables is the slope of a tangent line created when we intersect a vertical plane at afixed level of the other variable with the surface R3. Consider the function

f(x1, x2) = 10x1/41 x

1/22

which is shown in figure 22 along with a vertical plane at x2 = 10.

Now consider figure 23 which highlights the intersection of the vertical plane and the surface repre-senting the function. This line is a graph of the function f(x1, 10) = 10x

1/41

√10. Figure 24 shows this

intersection line alone. The partial derivative of f(x1,x2) with respect to x1 represents the slope of the tan-gent to this curve at a given point. Figure 25 shows the tangent to the curve representing the intersection ofthe vertical plane at x2 = 10 and the surface, while figure 26 shows all of the graphs together.


FIGURE 22. Function f(x1, x2) = 10x1/41 x

1/22 with Vertical Plane at x2 = 10

0

10

20x1

0

10

20x2

0

25

50

75

100

fHx1, x2L

0

10

20x1

0

10

20x2

FIGURE 23. Intersection of function f(x1, x2) = 10x1/41 x

1/22 and Vertical Plane at x2 = 10

0

5

10

15

20

x1

0

5

10

15

2025

x2

0

20

40

60

80

100

fHx1, x2L

0

5

10

15x1

0

5

10

15

20x2

Figure 27 shows the tangent to the curve representing the intersection of the vertical plane at x1 = 1 andthe surface f(x1, x2) = 1

2 (x1− 1)x2 − (x1− 1)2−x22. Figure 28 shows the tangent to the curve representing

the intersection of the vertical plane at x1 = 1 and the surface f(x1, x2) = (x1 − 2)x2 + x22 + ex1 x2 . Figure 29

shows tangents to the curve in both the x1 and x2 directions for the surface f(x1, x2) = 4e−x2

1 + x22

4 + x226 .

8. TANGENT LINES AND PLANES FOR FUNCTIONS WITH DOMAIN IN R2

The equation for the plane that is tangent to the graph of y = f(x1,x2) at the point (x01, x0

2, f(x01, x0

2)) is


FIGURE 24. The function f(x1, 10) = 10x1/41 101/2

0

5

10

15

20

0

5

10

1520

25

0

20

40

60

80

100

0

5

10

15

0

5

10

1520

FIGURE 25. Tangent to the function f(x1, x2) = 10x1/41 x

1/22 when x2 is fixed at 10 and x1 = 4

04

812

16x1

0

510

1520x2

0

20

40

60

80

fHx1, x2L

04

812

16x1

0

510

1520x2

y = f(x01, x0

2) +∂f(x0

1, x02)

∂x1

(x1 − x0

1

)+

∂f(x01, x0

2)∂x2

(x2 − x0

2

)(4)

Intuitively, this says we approximate the function f by its value at the point (x01, x0

2), then move away

from (x01, x0

2) along the plane that is tangent to the surface at that point. This plane has slope∂f(x0

1, x02)

∂x1in




0

5

10

15

20

x1

0

5

10

1520

25

x2

0

20

40

60

80

100

fHx1, x2L

0

5

10

15x1

0

5

10

1520x2

the x1 direction and slope∂f(x0

1, x02)

∂x2in the x2 direction. Equation 4 is also called a first order Taylor series

approximation to the function f at the point {x01, x0

2}.

A tangent plane contains the tangent lines when we hold x1 and x2 respectively constant. If we hold x2

constant at x02, then equation 4 reduces to

y = f(x01, x0

2) +∂f(x0

1, x02)

∂x1

(x1 − x0

1

)+

∂f(x01, x0

2)∂x2

(x0

2 − x02

)

= f(x01, x0

2) +∂f(x0

1, x02)

∂x1

(x1 − x0

1

) (5)

which is just the equation for a tangent line when the domain of the function is R1.

Consider figure 25 where we hold x2 constant at 10. This looks just like a tangent graph when the domainof the function is the real line. Then consider figure 30 where we hold x1 constant at 4 and vary x2. If wesimplify this graph as with figure 25, we obtain a graph similar to a tangent graph when the domain ofthe function is the real line and the independent variable is x2. This is shown in figure 31. If we rotate thisgraph as in figure 32, we can see clearly that it represents the value of f(x1, x2) as a function of x2.

The plane tangent to the surface f(x1, x2) = 10x1/41 x

1/22 at the point {x1=4, x2 =10 } is shown in figure 33.

The plane tangent to the surface f(x1, x2) = (x1 − 2)x2 + x22 + ex1 x2 is shown in figure 34.

9. HIGHER ORDER PARTIAL DERIVATIVES

9.1. Second order partial derivatives. When we differentiate a function f(x1, x2) twice, we produce itssecond order derivatives. These derivatives are usually denoted by


FIGURE 27. Tangent to the function f(x1, x2) = 12 (x1 − 1)x2 − (x1 − 1)2 − x2

2 when x1 isfixed at 1 and x2 = - 1

2

00.5

11.5

2

x1

-1-0.5

00.5

1

x1

-3

-2

-1

0

1

fHx1, x2L

-3

-2

-

0

∂2f∂ x2

1or fx1x1 or f11

∂2f∂ x2

2or fx2x2 or f22

∂2f∂ x1 ∂ x2

or fx1 x2 or f12

∂2f∂ x2 ∂ x1

or fx2 x1 or f21

(6)

The defining equations are

∂2f

∂x21

=∂

∂ x1

∂f

∂ x1

∂2f

∂x1∂x2=

∂

∂x1

∂f

∂ x2

(7)

Notice that the order in which the derivatives are taken, ∂2f∂ x1,∂ x2

means differentiate first with respect tox2, then with respect to x1.


FIGURE 28. Tangent to the function f(x1, x2) = (x1 − 2)x2 + x22 + ex1 x2 when x1 is fixed at

1 and x2 = 12

0.50.75

11.25

1.5

x1

-20

2 4x2

0

10

20

30

40

f Hx1,x2L

0.50.75

11.25x1

-20

2

Here are some example of first and second order partial derivatives.

a. Here is a function and the first and second order partial derivatives.

f (x, w) = 50x2w + 3wx

∂f

∂x= 100xw + 3w

∂f

∂w= 50x2 + 3x

∂2f

∂x2= 100w

∂2f

∂w∂x= 100x + 3

∂2f

∂x ∂w= 100x + 3

∂2f

∂ w2= 0

b. Here is a function and the first and second order partial derivatives.


FIGURE 29. Tangents to the function f(x1, x2) = 4e−x2

1 + x22

4 + x226 at (0.5, 0.3)

-20

2

-20

2

0

2

4

6

0

2

4

6

f (x1 x2, x3) = 50 + 5x1 + 3 x2 + 2 x3 + 7x21 + 2 x1x2 + 3x1 x3 + 5 x2

2 + 4 x2 x3 + 2x23

∂f

∂x1= 5 + 14x1 + 2x2 + 3x3

∂ f

∂ x2= 3 + 2x1 + 10x2 + 4x3

∂ f

∂ x3= 2 + 3x1 + 4x2 + 4x3

∂2f

∂ x21

= 14,∂2f

∂ x2 ∂x1= 2,

∂2 f

∂ x3 ∂ x1= 3

∂2f

∂ x1 ∂x2= 2,

∂2f

∂ x22

= 10,∂2f

∂ x3 ∂ x2= 4

∂2f

∂ x1 ∂ x3= 3,

∂2f

∂ x2 ∂x3= 4,

∂2f

∂ x23

= 4




0

5

10

15

20

x1

0

5

10

1520

25

x2

0

20

40

60

80

100

fHx1, x2L

0

5

10

15x1

0

5

10

1520x2

FIGURE 31. Tangent to the function f(4, x2) = 10√

2x1/22 when x2 = 10

04

812

16x1

0

5

1015

20x2

0

20

40

60

80

fHx1, x2L

04

812

16x1

0

5

1015

20x2

c. Here is another function and the first and second order partial derivatives.


FIGURE 32. Rotated graph of tangent to the function f(4, x2) = 10√

2x1/22 when x2 = 10

04

812

16

x1

0 5 10 15 20x2

0

20

40

60

80

fHx1, x2L

0

20

40

60

80

L

FIGURE 33. Plane tangent to the function f(x1, x2) = 10x1/41 x

1/22 at the point x1 = 4 and x2 = 10

05

1015

20

x1

0

5

10

1520

x2

0

20

40

60

80

100

fHx1, x2L

05

1015x1

0

5

10

15x2

f(x1, x2, x3) = 50x0.21 x0.3

2 x0.43

∂f

∂x1= 10x−0.8

1 x0.32 x0.4

3

∂f

∂x2= 15x0.2

1 x−0.72 x0.4

3

∂f

∂x3= 20x0.2

1 x0.32 x−0.6

3

∂2f

∂x21

= − 8x−1.81 x0.3

2 x0.43 ,

∂2f

∂x2∂x1= 3x−0.8

1 x−0.72 x0.4

3 ,∂2f

∂x3∂x1= 4x−0.8

1 x0.32 x−0.6

3

∂2f

∂x1∂x2= 3x−0.8

1 x−0.72 x0.4

3 ,∂2f

∂x22

= −10.5x0.21 x−1.7

2 x0.43 ,

∂2f

∂x3∂x2= 6x0.2

1 x−0.72 x−0.6

3

∂2f

∂x1∂x3= 4x−0.8

1 x0.32 x−0.6

3 ,∂2f

∂x2∂x3= 6x0.2

1 x−0.72 x−0.6

3 ,∂2f

∂x23

= −12x0.21 x0.3

2 x−1.63


FIGURE 34. Plane tangent to the function f(x1, x2) = (x1 − 2)x2 + x22 + ex1 x2 when x1 = 1

and x2 = 12

0.50.75

1

1.25

1.5

x1

-20

24

x2

0

5

10

15

fHx1,x2L

0.50.75

1

1.25x1

-20

2

9.2. Young’s theorem. As should be obvious from the examples, ∂2f∂x2 ∂x1

= ∂2f∂x1 ∂x2

. This is more generallystated as Young’s theorem. If f(x1, x2, ... , xn) and its partial derivatives f1, f2, . . . , fn,f11, f12, . . . , f1n, f21, fnn

are defined throughout an open region containing a point (x01, x0

2, x03, . . . , x0

n ) and are all continuous at(x0

1, x02, x0

3, . . . , x0n ) then ∂2f

∂xj∂xi= ∂2f

∂xi∂xj.

9.3. Higher order derivatives. We can compute higher order partial derivatives just as we computedhigher order derivatives by simply differentiating again. Here is an example. Consider the function

f(x1, x2) = 10x1/41 x

1/22

The first order partial derivatives are

∂f

∂x1=

52x−3/41 x

1/22

∂f

∂x2= 5x

1/41 x

−1/22

The second order partial derivatives are

∂2f

∂x21

= − 158

x−7/41 x

1/22

∂2f

∂x2∂x1=

54x−3/41 x

−1/22

∂2f

∂x1∂x2=

54x−3/41 x

−1/22

∂2f

∂x22

= − 52x

1/41 x

−3/22

The third order partial derivatives are


∂2f

∂x31

=10532

x−11/41 x

1/22

∂2f

∂x2∂x1∂x1= − 15

16x−7/41 x

−1/22

∂2f

∂x1∂x2∂x1= − 15

16x−7/41 x

−1/22

∂2f

∂x2∂x2∂x1= − 5

8x−3/41 x

−3/22

∂2f

∂x1∂x2∂x2= − 5

8x−3/41 x

−3/22

∂2f

∂x32

=154

x1/41 x

−5/22

10. DEFINITION OF THE GRADIENT AND HESSIAN OF A FUNCTION OF n VARIABLES

10.1. Gradient of f. The gradient of a function of n variables f(x1, x2, . . . , xn) is defined as follows.

∇ f(x) =(

∂f

∂x1

∂f

∂x2. . .

∂f

∂xn

)′

=

∂f∂x1

∂f∂x2

...∂f

∂xn

(8)

10.2. Hessian matrix of f. The Hessian matrix of a function of n variables f(x1, x2, . . . , xn) is the n × nmatrix of second order partial derivatives. The Hessian will have different values depending on the valueof the vector x at which we choose to evaluate the various second derivatives. It is symmetric due toYoung’s theorem and looks as follows where we evaluate at the vector x0.

∇2 f(x) =[

∂2f

∂xi∂xj

]i, j = 1, 2, . . . , n

=

∂2f(x0)∂x1 ∂x1

∂2f(x0)∂x1 ∂x2

. . .∂2f(x0)∂x1 ∂xn

∂2f(x0)∂x2 ∂x1

∂2f(x0)∂x2 ∂x2

. . .∂2f(x0)∂x2 ∂xn

......

......

∂2f(x0)∂xn ∂x1

∂2f(x0)∂xn ∂x2

. . .∂2f(x0)∂xn ∂xn

(9)


11. EVALUATING THE DOUBLE INTEGRAL

11.1. Definitions. Consider a closed and bounded set in R2 denoted by Q.We will assume that Q is a basicregion. A basic region is a connected set in which the total boundary consists of a finite number of continuousarcs of the form x2 = φ(x1), x1 = ψ(x2). Let f be a real-valued function f that is continuous on Q. We want todefine the double integral

∫∫

Q

f(x1, x2) dx1dx2 (10)

Geometrically we can think of the double integral as the volume of the solid that is bounded below byQ and above by f(x1,x2). To define the integral we enclose Q by a rectangle R with sides parallel to thecoordinate axis. We extend f to all of R by setting f equal to zero outside of Q. This extended function,whichwe continue to call f, is bounded on R, and is continuous on all of R except possibly at the boundary of Q.In such a case f is integrable on R, that is, there exists a unique number I such that

Lf (P ) ≤ I ≤ Uf (P ) (11)for all partitions P of R. We define Lf (P) and Uf (P) based on nonoverlapping rectangles that partition R.

If we partition x1 into m sections and x2 into n sections then we have m × n nonoverlapping rectangles Rij .

Rij : xi−11 ≤ x1 ≤ xi

1, xj−12 ≤ x2 ≤ xj

2 (12)where 1 ≤ i ≤ m, 1 ≤ j ≤ n. For each rectangle, Rij , f takes on a maximum value and minumum

value. It will take a zero value for points in a given rectangle that are outside of Q. We know it takes on amaximum and minimum value because f is continous and Rij is closed and bounded. We will denote themaximum value that f takes on Rij by Mij and the minimum value that f takes on Rij by mij . We thencompute area of the rectangular prism

(Bij

)with the rectangle Rij as its base and Mij as its height as

Area(Bij

)= M ij ×Area

(Rij

)= M ij

(xi

1 − xi−11

) (xj

2 − xj−12

)= M ij∆xi

1 ∆xj2 (13)

We then compute Uf (P) as

Uf (P ) =m∑

i=1

n∑

j=1

M ij∆xi1 ∆xj

2 (14)

and then compute Lf (P) as

Lf (P ) =m∑

i=1

n∑

j=1

mij∆xi1 ∆xj

2 (15)

Note that∫∫

Q

1 dx1dx2 =∫∫

Q

dx1dx2 (16)

gives the volume of solid of constant height 1 erected over Q or

area of Q =∫∫

Q

dx1dx2 (17)


11.2. Properties.(a) Linearity

∫∫

Q

[αf(x1, x2) + βg(x1, x2)] dx1dx2 = α

∫∫

Q

f(x1, x2)dx1dx2 + β

∫∫

Q

g(x1, x2)dx1dx2 (18)

(b) Order

if f ≥ 0 on Q then∫∫

Q

f(x1, x2)dx1dx2 ≥ 0

if f ≤ g on Q then∫∫

Q

f(x1, x2)dx1dx2 ≤∫∫

Q

g(x1, x2)dx1dx2

(19)

(c) AdditivityIf Q is broken up into a finite number of non-overlapping basic regions Q1, Q2, . . . , Qn, then

∫∫

Q

f(x1, x2)dx1dx2 =∫∫

Q1

f(x1, x2)dx1dx2 +∫∫

Q2

f(x1, x2)dx1dx2 + . . . +∫∫

Qn

f(x1, x2)dx1dx2 (20)

(d) Mean Value TheoremIf f and g are continous functions of a basic region and if g is non-negative on Q, then there exists

a point (x01,x0

2) in Q for which∫∫

Q

f(x1, x2)g(x1, x2)dx1dx2 = f(x01, x

02)

∫∫

Q

g(x1, x2)dx1dx2 (21)

For the case of g(x01,x0

2) = 1, we obtain∫∫

Q

f(x1, x2)dx1dx2 = f(x01, x

02)

∫∫

Q

dx1dx2

= f(x01, x

02)× area of Q

(22)

11.3. Evaluation of double intergrals by repeated integrals. Difficulty in evaluating a double integral∫∫

Q

f(x1, x2)dx1dx2

can come from two sources: the integrand f or from the base region Q. In this section we consider atechnique for evaluating double integrals of continuous functions of over regions of types we call Type Iand Type II.

11.3.1. Type I Regions. The projection of Q onto the x1 axis is a closed interval [a,b] and Q consists of allpoints (x1, x2) with

a ≤ x1 ≤ b and φ1(x1) ≤ x2 ≤ φ2(x1)Consider figure 35 which describes a region of Type I.For a region of Type I, we compute the double integral iteratively as


FIGURE 35. Region of Type I

a bx1

x2

x2 = Φ2Hx1L

x2 = Φ1Hx1L

Q

a bx1

x2

∫∫

Q

f(x1, x2)dx1dx2 =

b∫

a

φ2(x1)∫

φ1(x1)

f(x1, x2)dx2

dx1 (23)

We first calculate

φ2(x1)∫

φ1(x1)

f(x1, x2)dx2 (24)

by integrating f(x1,x2) with respect to x2 from x2 = φ1(x1) to x2 = φ2(x1). The resulting expression is afunction of x1 alone, which we then integrate with respect to x1 from x1 = a to x1 = b.

11.3.2. Type II Regions. The projection of Q onto the x2 axis is a closed interval [c,d] and Q consists of allpoints (x1, x2) with

c ≤ x2 ≤ d and ψ1(x2) ≤ x1 ≤ ψ2(x2)

Consider figure 36 which describes a region of Type II.For a region of Type II, we compute the double integral iteratively as

∫∫

Q

f(x1, x2)dx1dx2 =

d∫

c

ψ2(x2)∫

ψ1(x2)

f(x1, x2)dx1

dx2 (25)

We first calculate


FIGURE 36. Region of Type II

c

d

x1 = j1Hx2Lx1 = j2Hx2L Q

ψ2(x2)∫

ψ1(x2)

f(x1, x2)dx1 (26)

by integrating f(x1,x2) with respect to x1 from x1 = ψ1(x2) to x1 = ψ2(x2). The resulting expression is afunction of x2 alone, which we then integrate with respect to x2 from x2 = c to x1 = d.

11.4. Examples.

11.4.1. Evaluation over a rectangle. Consider the following integral

∫∫

Q

f(x1, x2)dx1dx2 =

3∫

0

2∫

1

x21x2 dx2dx1 (27)

First evaluate

2∫

1

x21x2 dx2 (28)

This will give


∫ 2

1

x21x2 dx2 =

(12x2

1x22

)|21

=[12

x21 22

]−

[12

x21 12

]

=[2 x2

1

] −[12x2

1

]

=32x2

1

(29)

Now integrate this function from 0 to 3 as follows

∫ 3

0

32x2

1 dx1 =(

x31

2

)|30

=[272

]−

[02

]

=272

(30)

Now consider the following integral

∫∫

Q

f(x1, x2)dx1dx2 =

2∫

1

3∫

0

x21x2 dx1dx2 (31)

First evaluate

3∫

0

x21x2 dx1 (32)

This will give

∫ 3

0

x21x2 dx1 =

(13x3

1x2

)|30

=[13

33x2

]−

[13

03x2

]

= [9x2] − 0

= 9x2

(33)

Now integrate this function from 1 to 2 as follows


∫ 2

1

9x2 dx2 =(

9x2

2

2

)|21

=[362

]−

[92

]

=272

(34)

11.4.2. Evaluation over a general region 1. Consider the following integral

∫∫

Q

f(x1, x2)dx1dx2 =

1∫

−1

x21+1∫

2x21

(x1 + 2x2) dx2 dx1 (35)

First evaluate

x21+1∫

2x21

(x1 + 2x2) dx2 (36)

This will give

x21+1∫

2x21

(x1 + 2x2) dx2 =(x1x2 + x2

2

) |x21+1

2x21

=(x1(x2

1 + 1) + (x21 + 1)2

) − (x1(2x2

1) + (2x21)

2)

=(x3

1 + x1 + x41 + 2x2

1 + 1) − (

2x31 + 4x4

1

)

= − 3x41 − x3

1 + 2x21 + x1 + 1

(37)

Now integrate this function from -1 to 1 as followsZ 1

−1

�−3x41 − x3

1 + 2x21 + x1 + 1

�dx1 =

�−3x5

1

5− x4

1

4+

2x31

3+

x21

2+ x1

�|1−1

=

�−3

5− 1

4+

2

3+

1

2+ 1

�−�

3

5− 1

4+−2

3+

1

2− 1

�=

�−36

60− 15

60+

40

60+

30

60+

60

60

�−�

36

60− 15

60+−40

60+

30

60− 60

60

�=−72

60+

80

60+

120

60

=128

60

=32

15(38)


Consider a graph of the function in equation 35 in figure 37 where the bounds for integration for x1 are− 1

2 and 12 .

FIGURE 37. Integral as area under a curve over a region

-2

-1

0

1

2

x1

-2

0

2

4

x2

-5

0

5

fHx1,x2L-2

-1

0

1

2

x1

0

5

SIMPLE MULTIVARIATE CALCULUS - Economics · SIMPLE MULTIVARIATE CALCULUS 1. ... A real-valued...

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