Chapter 5 Partial Derivatives

Chapter 5

Partial Derivatives

5.1 Functions of Several Variables

The functions discussed in previous chapter involved only one independent variable.Such functions have many applications; however, in some problems several independentvariables occur. For instance, the ideal gas law P = ρRT states that the pressure P is afunction of both its density ρ and its temperature T . (The gas constant R is a materialproperty and not a variable).

In this section, we expand our concept of function to include functions that dependon more than one variable, that is, functions whose domain is multi-dimensional.

The terminology and notation for functions of several variables is similar to that forfunctions of one variable. For example, the expression

z = f(x, y)

means that z is a function of x and y in the sense that a unique value of the dependentvariable z is determined by specifying values for the independent variables x and y.Similarly,

w = f(x, y, z)

expresses w as a function x, y, and z, and

u = f(x1, x2, . . . , xn)

expresses u as a function of x1, x2, . . . , xn.As with functions of one variable, the independent variables of a function of two or

more variables may be restricted to lie in some set D, which we call the domain of f .

A function f of two variables, x and y, is a rule that assigns a uniquereal number f(x, y) to each point (x, y) in some set D in the xy-plane.

Definition 5.1

129

MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 130

A function f of three variables, x, y, and z, is a rule that assigns a uniquereal number f(x, y, z) to each point (x, y, z) in some set D in three dimensionalspace.

Definition 5.2

Let f(x, y) = 3x2√y − 1. Find f(1, 4), f(t2, t), f(ab, 9b), and the domain of f .

Example 5.1

Solution

Find and sketch the domain for (a) f(x, y) = ln(x2−y) and (b) g(x, y) =2x

y − x2.

Example 5.2

Solution


Let f(x, y, z) =√

1− x2 − y2 − z2. Find f(

0, 12,−1

2

)

and the domain of f .

Example 5.3

Solution

Graphs of Functions of Two Variables

Recall that for a function f of one variable, the graph of f(x) in the xy-plane was definedto be the graph of the equation y = f(x). Similarly, if f is a function of two variables,we define the graph of f(x, y) in xyz-space to be the graph of the equation z = f(x, y).In general, such a graph will be a surface in 3-space.

In each part, describe the graph of the function in an xyz-coordinate system.

(a) f(x, y) = 1− x− 12y (b) f(x, y) =

√

1− x2 − y2

(c) f(x, y) = −√

x2 + y2

Example 5.4

Solution


Exercise 5.1

1− 8 These exercises are concerned with functions of two variables.

1. Let f(x, y) = x2y + 1. Find

(a) f(2, 1) (b) f(1, 2) (c) f(0, 0)

(d) f(1,−3) (e) f(3a, a) (f) f(ab, a− b)

2. Let f(x, y) = x+ 3√xy. Find

(a) f(t, t2) (b) f(x, x2) (c) f(2y2, 4y)

3. Let f(x, y) = xy + 3. Find

(a) f(x+ y, x− y) (b) f(xy, 3x2y3)

4. Let g(x, y) = x sin y. Find

(a) g(x/y) (b) g(xy) (c) g(x− y)

5. Find F (g(x), h(x)) if F (x, y) = xexy, g(x) = x3, and h(y) = 3y + 1.

6. Find g(u(x, y), v(x, y)) if g(x, y) = y sin(x2y), u(x, y) = x2y3, and v(x, y) = πxy.

7. Let f(x, y) = x+ 3x2y2, x(t) = t2, and y(t) = t3. Find

(a) f(x(t), y(t)) (b) f(x(0), y(0)) (c) f(x(2), y(2))

8. Let g(x, y) = ye−3x, x(t) = ln(t2 + 1), and y(t) =√t. Find g(x(t), y(t)).

9− 12 These exercises involve functions of three variables.

9. Let f(x, y, z) = xy2z3 + 3. Find

(a) f(2, 1, 2) (b) f(−3, 2, 1)

(c) f(0, 0, 0) (d) f(a, a, a)

10. Let f(x, y, z) = zxy + x. Find

(a) f(x+ y, x− y, x2) (b) f(xy, y/x, xz)

11. Find F (f(x), g(y), h(z)) if F (x, y, z) = yexyz, f(x) = x2, g(y) = y + 1, and h(z) =z2.

12. Find g(u(x, y, z), v(x, y, z), w(x, y, z)) if g(x, y, z) = z sin xy, u(x, y, z) = x2z3,v(x, y, z) = πxyz, and w(x, y, z) = xy/z.


13− 14 These exercises are concerned with functions of four or more variables.

13. (a) Let f(x, y, z, t) = x2y3√z + t. Find f(

√5, 2, π, 3π).

(b) Let f(x1, x2, . . . , xn) =

n∑

k=1

kxk. Find f(1, 1, . . . , 1).

14. (a) Let f(u, v, λ, φ) = eu+v cosλ tanφ. Find f(−2, 2, 0, π/4).

(b) Let f(x1, x2, . . . , xn) = x21 + x2

2 + · · ·+ x2n. Find f(1, 2, . . . , n).

15− 18 Sketch the domain of f . Use solid lines for portions of the boundaryincluded in the domain and dashed lines for portions not included.

15. f(x, y) = ln(1− x2 − y2) 16. f(x, y) =√

x2 + y2 − 4

17. f(x, y) =1

x− y218. f(x, y) = ln xy

19− 20 Describe the domain of f in words.

19. (a) f(x, y) = xe−√y+2 (b) f(x, y, z) =

√

25− x2 − y2 − z2

(c) f(x, y, z) = exyz

20. (a) f(x, y) =

√4− x2

y2 + 3(b) f(x, y) = ln(y − 2x)

(c) f(x, y, z) =xyz

x+ y + z

21− 30 Sketch the graph of f .

21. f(x, y) = 3 22. f(x, y) =√

9− x2 − y2

23. f(x, y) =√

x2 + y2 24. f(x, y) = x2 + y2

25. f(x, y) = x2 − y2 26. f(x, y) = 4− x2 − y2

27. f(x, y) =√

x2 + y2 + 1 28. f(x, y) =√

x2 + y2 − 1

29. f(x, y) = y + 1 30. f(x, y) = x2

Answers to Exercise 5.1

1. (a) 5 (b) 3 (c) 1 (d) −2 (e) 9a3 + 1 (f) a3b2 − a2b3 + 1

3. (a) x2 − y2 + 3 (b) 3x3y4 + 3 5. x3ex3(3y+1) 7. (a) t2 + 3t10 (b) 0 (c) 3076

9. (a) 19 (b) −9 (c) 4 (d) a6 + 3 (e) −t8 + 3 (f) (a+ b)(a− b)2b3 + 3

11. (y + 1)ex2(y+1)z2 13. (a) 80

√π (b) n(n + 1)/2


15.

x

y

1

17.

x

y

19. (a) all points above or on the line y = −2

(b) all points on or within the sphere x2 + y2 + z2 = 25

(c) all points in 3-space

5.2 Limits and Continuity

For a function of a single variable, if we write

limx→a

f(x) = L,

we mean that x gets closer and closer to a, f(x) gets closer and closer to the numberL. Recall that when we say that x gets closer and closer to a, we mean that x getsarbitrarily close to a and can approach a from either side of a (x < a or x > a). Further,the limit must be the same as x approaches a from either side.

For functions of several variables, the idea is very similar. When we write

lim(x,y)→(a,b)

f(x, y) = L,

we mean that as (x, y) gets closer and closer to (a, b), f(x, y) is getting closer and closerto the number L. In this case, (x, y) may approach (a, b) along any path through. Notethat unlike the case for functions of a single variable, there are many (in fact, infinitelymany) different paths passing through any given point (a, b).


x

y

b(a, b)

b (x, y)

For instance,lim

(x,y)→(2,3)(x2 − xy + y2 − 2)

asks us to identify what happens to the function x2 − xy+ y2 − 2 as x approaches 2 andy approaches 3. Clearly, x2−xy+ y2− 2 approaches 22− 2(3)+ 32− 2 = 5 and we write

lim(x,y)→(2,3)

(x2 − xy + y2 − 2) = 5.

Similarly, we can reason that

lim(x,y)→(1,−π)

(sin xy + x2y) = sin(−π) + (−π) = −π

In other words, for many functions, we can compute limits simply by substituting in tothe function.

Unfortunately, as with functions of a single variables, the limits we’re most interestedin cannot computed by simply substituting values for x and y. For instance, for

lim(x,y)→(3,1)

y

x− 2y − 1

substituting in x = 3 and y = 1 gives the indeterminate form 00. To evaluate this limit,

we must investigate further.

Let f be defined on the interior of a circle center at the point (a, b), exceptpossibly at (a, b) itself. We say that

lim(x,y)→(a,b)

f(x, y) = L,

if for every ǫ > 0 there exists a δ > 0 such that |f(x, y)− L| < ǫ whenever

0 <√

(x− a)2 + (y − b)2 < δ.

Definition 5.3

The basic properties of limits of functions of one variable discussed in Calculus I areextended to the case of functions of several variables.


Let c be a number. Suppose that

lim(x,y)→(a,b)

f(x, y) = L1 and lim(x,y)→(a,b)

g(x, y) = L2.

Then the following properties hold:

1. lim(x,y)→(a,b)

c = c

2. lim(x,y)→(a,b)

x = a and lim(x,y)→(a,b)

y = b

3. lim(x,y)→(a,b)

cf(x, y) = c lim(x,y)→(a,b)

f(x, y) = cL1

4. lim(x,y)→(a,b)

[f(x, y)± g(x, y)] = lim(x,y)→(a,b)

f(x, y)± lim(x,y)→(a,b)

g(x, y) = L1±L2

5. lim(x,y)→(a,b)

f(x, y) · g(x, y) = lim(x,y)→(a,b)

f(x, y) · lim(x,y)→(a,b)

g(x, y) = L1 · L2

6. lim(x,y)→(a,b)

f(x, y)

g(x, y)=

lim(x,y)→(a,b)

f(x, y)

lim(x,y)→(a,b)

g(x, y)=

L1

L2, if L2 6= 0

Theorem 5.1 (Properties of the Limit)

Remark: As for single-variable functions, limits of polynomial and rational functionsin two variables may be found by substituting for x and y.

Evaluate lim(x,y)→(2,1)

5xy2 + 3y

2x2y + 3xy.

Example 5.5

Solution



x3 − y3

x− y.

Example 5.6

Solution


x2 − xy√x−√

y.

Example 5.7

Solution


Remark: If f(x, y) approaches L1 as (x, y) approaches (a, b) along a path P1 and f(x, y)approaches L2 6= L1 as (x, y) approaches (a, b) along a path P2, then lim

(x,y)→(a,b)f(x, y)

does not exist.

Show that lim(x,y)→(0,0)

x2y2

x4 + 3y4does not exist.

Example 5.8

Solution

If C is a smooth parametric curve in 2-space or 3-space that is represented by theequations

x = x(t), y = y(t) or x = x(t), y = y(t), z = z(t)

and if x0 = x(t0), y0 = y(t0), and z0 = z(t0), then the limits

lim(x, y) → (x0, y0)

(along C)

f(x, y) and lim(x, y, z) → (x0, y0, z0)

(along C)

f(x, y, z)

are defined by

lim(x, y) → (x0, y0)

(along C)

f(x, y) = limt→t0

f(x(t), y(t)) (5.1)

lim(x, y, z) → (x0, y0, z0)

(along C)

f(x, y, z) = limt→t0

f(x(t), y(t), z(t)) (5.2)


Find the limit of the function f(x, y) = − xy

x2 + y2as (x, y) → (0, 0) along

(a) the x-axis (b) the line y = x (c) the parabola y = x2

Example 5.9

Solution

Relationships between General Limits and Limit Along Smooth

Curve

(a) If f(x, y) → L as (x, y) → (a, b), then f(x, y) → L as (x, y) → (a, b) alongany smooth curve.

(b) If the limit of f(x, y) fail to exit as (x, y) → (a, b) along some smoothcurve, or if f(x, y) has different limit as (x, y) → (a, b) along two differentsmooth curves, then the limit of f(x, y) does not exist as (x, y) → (a.b).

Theorem 5.2


The limitlim

(x,y)→(0,0)− xy

x2 + y2

does not exist because in Example 5.9 we found two different smooth curvesalong which this limit had different values. Specifically,

lim(x, y) → (0, 0)(along y = 0)

− xy

x2 + y2= 0

and

lim(x, y) → (0, 0)(along y = x)

− xy

x2 + y2= −1

2. z

Example 5.10

Continuity

A function f(x, y) is said to be continuous at (a, b) if f(a, b) is defined and if

lim(x,y)→(a,b)

f(x, y) = f(a, b)

In addition, if f is continuous at every point in an open set D, then we say thatf is continuous on D, and if f is continuous at every point in the xy-plane,then we say that f is continuous everywhere.

Definition 5.4

Let f(x, y) =xy

x2 + y2. Determine whether the function f is continuous at (1, 0).

Example 5.11

Solution


Let f(x, y) =

x2y2

x4 + 3y4if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0)Determine whether the function f is continuous at (0, 0).

Example 5.12

Solution

Let

f(x, y) =

x2 − 2xy − 3y2

x− 3yif x− 3y 6= 0

2 if x− 3y = 0

Determine whether the function f is continuous at (3, 1).

Example 5.13

Solution


Recognizing continuous Functions

• A composition of continuous functions is continuous.

• A sum, difference, or product of continuous functions is continuous.

• A quotient of continuous functions is continuous, except where the denominator iszero.

Since the function

f(x, y) =x3y2

1− xy

is a quotient of continuous functions, it is continuous except where 1− xy = 0.Thus f(x, y) is continuous everywhere except on the hyperbola xy = 1. z

Example 5.14

(a) If g(x) is continuous at a and h(y) is continuous at b, then f(x, y) =g(x)h(y) is continuous at (a, b).

(b) If h(x, y) is continuous at (a, b) and g(u) is continuous at u = h(a, b), thenthe composition f(x, y) = g(h(x, y)) is continuous at (a, b).

(c) If f(x, y) is continuous at (a, b) and if x(t) and y(t) is continuous at t0 withx(t0) = a and y(t0) = b, then the composition f(x(t), y(t)) is continuousat t0.

Theorem 5.3

Show that the function f(x, y) = ex2−3xy+2y2 is continuous everywhere.

Example 5.15

Solution If h(x, y) = x2 − 3xy + 2y2 and g(u) = eu, then f(x, y) = g(h(x, y)).Since g is continuous for all values of u and h is a polynomial in x and y (and hence con-tinuous for all x and y), it follows that the composition f(x, y) = g(h(x, y)) is continuouseverywhere. z

Exercise 5.2

1− 6 Use limit laws and continuity properties to evaluate the limit.

1. lim(x,y)→(1,3)

(4xy2 − x) 2. lim(x,y)→(1/2,π)

(xy2 sin xy)

3. lim(x,y)→(−1,2)

xy3

x+ y4. lim

(x,y)→(1,−3)e2x−y2

5. lim(x,y)→(0,0)

ln(1 + x2y3) 6. lim(x,y)→(4,−2)

x 3

√

y3 + 2x


7− 10 Show that the limit does not exist by considering the limits as(x, y) → (0, 0) along the coordinate axes.

7. lim(x,y)→(0,0)

3

x2 + 2y28. lim

(x,y)→(0,0)

x+ y

2x2 + y2

9. lim(x,y)→(0,0)

x− y

x2 + y210. lim

(x,y)→(0,0)

cosxy

x2 + y2

11− 14 Evaluate the limit using the substitution z = x2 + y2 and observe thatz → 0+ if and only if (x, y) → (0, 0).

11. lim(x,y)→(0,0)

sin(x2 + y2)

x2 + y212. lim

(x,y)→(0,0)

1− cos(x2 + y2)

x2 + y2

13. lim(x,y)→(0,0)

e−1/(x2+y2) 14. lim(x,y)→(0,0)

e−1/√

x2+y2

√

x2 + y2

15− 22 Determine whether the limit exists. If so, find its value.

15. lim(x,y)→(0,0)

x4 − y4

x2 + y216. lim

(x,y)→(0,0)

x4 − 16y4

x2 + 4y2

17. lim(x,y)→(0,0)

xy

3x2 + 2y218. lim

(x,y)→(0,0)

1− x2 − y2

x2 + y2

19. lim(x,y,z)→(2,−1,2)

xz2√

x2 + y2 + z220. lim

(x,y,z)→(2,0,−1)ln(2x+ y − z)

21. lim(x,y,z)→(0,0,0)

sin(x2 + y2 + z2)√

x2 + y2 + z222. lim

(x,y,z)→(0,0,0)

sin√

x2 + y2 + z2

x2 + y2 + z2


1. 35 3. −8 5. 0 7. along x = 0, limit does not exist

8. along x = 0, limit does not exist

11. 1 13. 0 15. 0 17. limit does not exist 19. 83

21. 0

5.3 Partial Derivatives

Partial Derivatives of Functions of Two Variables

Recall that given a function of one variable, f(x) , the derivative, f ′(x), represents therate of change of the function as x changes. This is an important interpretation ofderivatives and we are not going to want to lose it with functions of more than onevariable.

In other words, what do we do if we only want one of the variables to change, or ifwe want more than one of them to change?

In this section we are going to concentrate exclusively on only changing one of thevariables at a time, while the remaining variable(s) are held fixed.


If z = f(x, y) and (a, b) is a point in the domain of f , then the partial deriva-tive of f with respect to x at (a, b) is the derivative at a of the function thatresults when y = b is held fixed and x is allowed to vary. This partial derivativeis denoted by fx(a, b) and is given by

fx(a, b) =d

dx[f(x, b)]

∣

∣

∣

x=a

= limh→0

f(a+ h, b)− f(a, b)

h(5.3)

Similarly, the partial derivative of f with respect to y at (a, b) is thederivative at b of the function that results when x = a is held fixed and y isallowed to vary. This partial derivative is denoted by fy(a, b) and is given by

fy(a, b) =d

dy[f(a, y)]

∣

∣

∣

y=b

= limh→0

f(a, b+ h)− f(a, b)

h(5.4)

Definition 5.5

Find fx(1, 0) and fy(2,−1) for the function f(x, y) = 3x2 + x3y + 4y2.

Example 5.16

Solution


The Partial Derivative Functions

Formulas (5.3) and (5.4) define the partial derivatives of a function at a specific point(a, b). However, often it will be desirable to omit the subscripts and think of the partialderivatives as a functions of the variables x and y. These functions are

fx(x, y) = limh→0

f(x+ h, y)− f(x, y)

h

and

fy(x, y) = limh→0

f(x, y + h)− f(x, y)

h

The following example gives an alternative way of performing the computations inExample 5.16.

Find fx(x, y) and fy(x, y) for f(x, y) = 3x2 + x3y + 4y2, and use those partialderivatives to compute fx(1, 0) and fy(2,−1).

Example 5.17

Solution

Partial Derivative Notation

If z = f(x, y), then the partial derivatives fx and fy are also denoted by the symbols

∂f

∂x,

∂z

∂xand

∂f

∂y,

∂z

∂y

Some typical notations for the partial derivatives of z = f(x, y) at a point (a, b) are

∂f

∂x

∣

∣

∣

∣

x=a,y=b

,∂z

∂x

∣

∣

∣

∣

(a,b)

,∂f

∂x

∣

∣

∣

∣

(a,b)

,∂f

∂x(a, b),

∂z

∂x(a, b)


Find∂z

∂xand

∂z

∂yif z = exy +

x

y.

Example 5.18

Solution

Partial Derivatives Viewed as Rates of Change and Slopes

Recall that if y = f(x), then the value of f ′(a) can be interpreted either as the rate ofchange of y with respect to x at a or as the slope of the tangent line to the graph of fat a. Partial derivatives have analogous interpretations.

Suppose that C1 is the intersection of the surface z = f(x, y) with the plane y = band that C2 is its intersection with the plane x = a.

z

x

y

b

C2

C1

z = f(x, y)

Thus, fx(x, b) can be interpreted as the rate of change of z with respect to x along thecurve C1 and fy(a, y) can be interpreted as the rate of change of z with respect to yalong the curve C2.

In particular, fx(a, b) is the rate of change of z with respect to x along the curve C1

at the point (a, b), and fy(a, b) is the rate of change of z with respect to y along thecurve C2 at the point (a, b).


For a real gas, van der Waals’s equation states that

(

P +n2a

V 2

)

(V − nb) = nRT

Here, P is the pressure of the gas, V is the volume of the gas, T is the temper-ature (in degrees Kelvin), n is the number of moles of gas, R is the universalgas constant and a and b are constants. Compute and interpret PV and TP

Example 5.19

Solution

Geometrically, fx(a, b) can be viewed as the slope of the tangent line to the curveC1 at the point (a, b), and fy(a, b) can be viewed as the slope of the tangent line to thecurve C2 at the point (a, b).

We will call fx(a, b) the slope of the surface in the x-direction at (a, b) andfy(a, b) the slope of the surface in the y-direction at (a, b).


Let f(x, y) = x2y + 5y3.

(a) Find the slope of the surface z = f(x, y) in the x-direction at the point(1,−2).

(b) Find the slope of the surface z = f(x, y) in the y-direction at the point(1,−2).

Example 5.20

Solution

Implicit Partial Differentiation

Find the slope of the sphere x2 + y2 + z2 = 1 in the y-direction at the point(

23, 13, 23

)

.

Example 5.21

Solution


Suppose that D =√

x2 + y2 is the length of the diagonal of a rectangle whosesides have lengths x and y that are allowed to vary. Find the formula for therate of change of D with respect to x if x varies with y held constant, and usethis formula to find the rate of change of D with respect to x at the point wherex = 3 and y = 4.

Example 5.22

Solution

Partial Derivatives of Functions with More Than Two Variables

For a function f(x, y, z) of three variables, there are three partial derivatives :

fx(x, y, z), fy(x, y, z), fz(x, y, z)

The partial derivative fx is calculated by holding y and z constant and differentiatingwith respect to x. For fy the variables x and z are held constant, and for fz the variablesx and y are held constant. If a dependent variable

w = f(x, y, z)

is used, then the three partial derivatives of f can be denoted by

∂w

∂x,

∂w

∂y, and

∂w

∂z


If f(x, y, z) = x3y2z4 + 2xy + z, then

fx(x, y, z) =

fy(x, y, z) =

fz(x, y, z) =

fz(−1, 1, 2) =

Example 5.23

If f(ρ, θ, φ) = ρ2 cos φ sin θ, then

fρ(ρ, θ, φ) =

fθ(ρ, θ, φ) =

fφ(ρ, θ, φ) =

Example 5.24

In general, if f(v1, v2, . . . , vn) is a function of n variables, there are n partial deriva-tives of f , each of which is obtained by holding n − 1 of the variables fixed and differ-entiating the function f with respect to the remaining variable. If w = f(v1, v2, . . . , vn),then these partial derivatives are denoted by

∂w

∂v1,∂w

∂v2, . . . ,

∂w

∂vn

where ∂w/∂vi is obtained by holding all variables except vi fixed and differentiating withrespect to vi.

Find∂

∂xi

[

√

x21 + x2

2 + · · ·+ x2n

]

for i = 1, 2, . . . , n.

Example 5.25

Solution For each i = 1, 2, . . . , n we obtain

∂

∂xi

[

√

x21 + x2

2 + · · ·+ x2n

]

=1

2√

x21 + x2

2 + · · ·+ x2n

· ∂

∂xi

[

x21 + x2

2 + · · ·+ x2n

]

=1

2√

x21 + x2

2 + · · ·+ x2n

[2xi]

=xi

√

x21 + x2

2 + · · ·+ x2n

z


Higher-Order Partial Derivatives

Suppose that f is a function of two variables x and y. Since the partial derivatives ∂f/∂xand ∂f/∂y are also functions of x and y, these functions may themselves have partialderivatives. This gives rise to four possible second-order partial derivatives of f , whichare defined by

∂2f

∂x2=

∂

∂x

(

∂f

∂x

)

= fxx∂2f

∂y2=

∂

∂y

(

∂f

∂y

)

= fyy

Differentiate twicewith respect to x.

Differentiate twicewith respect to y.

∂2f

∂y∂x=

∂

∂y

(

∂f

∂x

)

= fxy∂2f

∂x∂y=

∂

∂x

(

∂f

∂y

)

= fyx

Differentiate first withrespect to x and thenwith respect to y.

Differentiate first withrespect to y and thenwith respect to x.

The last two cases are called the mixed second-order partial derivatives or themixed second partials. Also, the derivatives ∂f/∂x and ∂f/∂x are often called thefirst-order partial derivatives when it is necessary to distinguish them from higher-order partial derivatives.

Find the second-order partial derivatives of f(x, y) = x2y − y3 + ln x.

Example 5.26

Solution


Let f be a function of two variables. If fxy and fyx are continuous on some opendisk, then fxy = fyx on that disk.

Theorem 5.4

Third-order, fourth-order, and higher-order partial derivatives can be obtained bysuccessive differentiation. Some possibilities are

∂3f

∂x3=

∂

∂x

(

∂2f

∂x2

)

= fxxx

∂4f

∂y4=

∂

∂y

(

∂3f

∂y3

)

= fyyyy

∂3f

∂y2∂x=

∂

∂y

(

∂2f

∂y∂x

)

= fxyy

∂4f

∂y2∂x2=

∂

∂y

(

∂3f

∂y∂x2

)

= fxxyy

Let f(x, y) = cos(xy)− x3 + y4. Find fxyy.

Example 5.27

Solution

Differentiability

A function f of two variables is said to be differentiable at (a, b) providedfx(a, b) and fy(a, b) both exist and

lim(∆x,∆y)→(0,0)

∆f − fx(a, b)∆x− fy(a, b)∆y√

(∆x)2 + (∆y)2= 0

Definition 5.6


Note For a function f(x, y), the symbol ∆f , called the increment of f , denotes thechange in the value of f(x, y) that results when (x, y) varies from some initial position(a, b) to some new position (a+∆x, b+∆y); thus

∆f = f(a+∆x, b+∆y)− f(a, b)

A function f of three variables is said to be differentiable at (a, b, c) providedfx(a, b, c), fy(a, b, c), and fz(a, b, c) exist and

lim(∆x,∆y,∆z)→(0,0,0)

∆f − fx(a, b, c)∆x− fy(a, b, c)∆y − fz(a, b.c)∆z√

(∆x)2 + (∆y)2 + (∆z)2= 0

Definition 5.7

If a function f of two variables is differentiable at each point of the region R in thexy-plane, then we say that f is differentiable on R; and if f is differentiable at everypoint in the xy-plane, then we say that f is differentiable everywhere.

For a function f of three variables we have corresponding conventions.

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

Theorem 5.5

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

Theorem 5.6

For example, consider the function

f(x, y, z) = x+ yz

Since fx(x, y, z) = 1, fy(x, y, z) = z, fz(x, y, z) = y are defined and continuous every-where, we conclude that f is differentiable everywhere.

Differentials

If z = f(x, y) is differentiable at a point (a, b), we let

dz = fx(a, b)dx+ fy(a, b)dy (5.5)

denote a new function with dependent variable dz and independent variables dx and dy.We refer to this function (also denoted df) as the total differential of z at (a, b) oras the total differential of f at (a, b).


Similarly, for a function w = f(x, y, z) of three variables we have the total differ-

ential of w at (a, b, c),

dw = fx(a, b, c)dx+ fy(a, b, c)dy + fz(a, b, c)dz (5.6)

which is also referred to as the total differential of f at (a, b, c). It is common practiceto omit the subscripts and write Equations (5.5) and (5.6) as

dz = fx(x, y)dx+ fy(x, y)dy (5.7)

anddw = fx(x, y, z)dx+ fy(x, y, z)dy + fz(x, y, z)dz (5.8)

Local Linear Approximations

We now show that if a function f is differentiable at a point, then it can be very closelyapproximated by a linear function near that point.

The linearization of a function f(x, y) at a point (a, b) where f is differentiableis the function

L(x, y) = f(a, b) + fx(a, b)(x− a) + fy(a, b)(y − b).

The approximationf(x, y) ≈ L(x, y)

is the local linear approximation to f at (a, b).

Definition 5.8

Find the linearization of f(x, y) = xexy at the point (1, 0). Use it to estimatef(1.1,−0.1).

Example 5.28

Solution


For a function f(x, y, z) that is differentiable at (a, b, c), the local linear approximationis

L(x, y, z) = f(a, b, c) + fx(a, b, c)(x− a) + fy(a, b, c)(y − b) + fz(a, b, c)(z − c).

Exercise 5.3

1. Let f(x, y) = 3x3y2. Find

(a) fx(x, y) (b) fy(x, y) (c) fx(1, y)

(d) fx(x, 1) (e) fy(1, y) (f) fy(x, 1)

(g) fx(1, 2) (h) fy(1, 2)

2. Let z = e2x sin y. Find

(a) ∂z/∂x (b) ∂z/∂y (c) ∂z/∂x|(0,y)(d) ∂z/∂x|(x,0) (e) ∂z/∂y|(0,y) (f) ∂z/∂y|(x,0)(g) ∂z/∂x|(ln 2,0) (h) ∂z/∂y|(ln 2,0)

3. Let f(x, y) =√3x+ 2y. Find

(a) Find the slope of the surface z = f(x, y) in the x-direction at the point (4, 2).

(b) Find the slope of the surface z = f(x, y) in the y-direction at the point (4, 2).

4. Let f(x, y) = xe−y + 5y. Find

(a) Find the slope of the surface z = f(x, y) in the x-direction at the point (3, 0).

(b) Find the slope of the surface z = f(x, y) in the y-direction at the point (3, 0).

5. Let z = sin(y2 − 4x). Find

(a) Find the rate of change of z with respect to x at the point (2, 1) with y heldfixed.

(b) Find the rate of change of z with respect to y at the point (2, 1) with x heldfixed.

6. Let z = (x+ y)−1. Find

(a) Find the rate of change of z with respect to x at the point (−2, 4) with y heldfixed.

(b) Find the rate of change of z with respect to y at the point (−2, 4) with x heldfixed.

7− 12 Find ∂z/∂x and ∂z/∂y.

7. z = 4ex2y3 8. z = cos(x5y4)

9. z = x3 ln(1 + xy−3/5) 10. z = exy sin 4y2

11. z =xy

x2 + y212. z =

x2y3√x+ y


13− 17 Find fx(x, y) and fy(x, y).

13. f(x, y) =√

5x5y − 7x3y 14. f(x, y) =x+ y

x− y

15. f(x, y) = y−3/2 tan−1(x/y) 16. f(x, y) = x3e−y + y3 sec√x

17. f(x, y) = (y2 tanx)−4/3

18− 21 Evaluate the indicated partial derivatives.

18. f(x, y) = 9− x2 − 7y3; fx(3, 1), fy(3, 1)

19. f(x, y) = x2yexy; ∂f/∂x(1, 1), ∂f/∂y(1, 1)

20. z =√

x2 + 4y2; ∂z/∂x(1, 2), ∂z/∂y(1, 2)

21. w = x2 cosxy; ∂w/∂x(

12, π

)

, ∂w/∂y(

12, π

)

22. Let f(x, y, z) = x2y4z3 + xy + z2 + 1. Find

(a) fx(x, y, z) (b) fy(x, y, z) (c) fz(x, y, z)

(d) fx(1, y, z) (e) fy(1, 2, z) (f) fz(1, 2, 3)

23. Let w = x2y cos z. Find

(a) ∂w/∂x(x, y, z) (b) ∂w/∂y(x, y, z) (c) ∂w/∂z(x, y, z)

(d) ∂w/∂x(2, y, z) (e) ∂w/∂y(2, 1, z) (f) ∂w/∂z(2, 1, 0)

24− 26 Find fx, fy, and fz.

24. f(x, y, z) = z ln(x2y cos z) 25. f(x, y, z) = y−3/2 sec

(

xz

y

)

26. f(x, y, z) = tan−1

(

1

xy2z3

)

27− 30 Find ∂w/∂x, ∂w/∂y, and ∂w/∂z.

27. w = yez sin xz 28. w =x2 − y2

y2 + z2

29. w =√

x2 + y2 + z2 30. w = y3e2x+3z

31. Let f(x, y, z) = y2exz. Find

(a) ∂f/∂x|(1,1,1) (b) ∂f/∂y|(1,1,1) (c) ∂f/∂z|(1,1,1)

32. Let w =√

x2 + 4y2 − z2. Find

(a) ∂w/∂x|(2,1,−1) (b) ∂w/∂y|(2,1,−1) (c) ∂w/∂z|(2,1,−1)

33. The volume V of a right circular cylinder is given by the formula V = πr2h, wherer is the radius and h is the height.


(a) Find the formula for the instantaneous rate of change of V with respect to rif r changes and h remains constant.

(b) Find the formula for the instantaneous rate of change of V with respect to hif h changes and r remains constant.

(c) Suppose that h has a constant value of 4 in, but r varies. Find the rate ofchange of V with respect to r at the point where r = 6 in.

(d) Suppose that r has a constant value of 8 in, but h varies. Find the instan-taneous rate of change of V with respect to h at the point where h = 10in.

34. The volume V of a right circular cone is given by

V =π

24d2√4s2 − d2

where s is the slant height and d is the diameter of the base.

(a) Find the formula for the instantaneous rate of change of V with respect to sif d remains constant.

(b) Find the formula for the instantaneous rate of change of V with respect to dif s remains constant.

(c) Suppose that d has a constant value of 16 cm, but s varies. Find the rate ofchange of V with respect to s when s = 10 cm.

(d) Suppose that s has a constant value of 10 cm, but d varies. Find the rate ofchange of V with respect to d when d = 16 cm.

35. According to the ideal gas law, the pressure, temperature, and volume of a gas arerelated by P = kT/V , where k is a constant of proprotionality. Suppose that V ismeasured in cubic inches (in3), T is measured in Kelvin (K), and that for a certaingas the constant of proprotionality is k = 10 in-lb/K.

(a) Find the instantaneous rate of change of pressure with respect to temperatureif the temperature is 80 K and the volume remain fixed at 50 in3.

(b) Find the instantaneous rate of change of volume with respect to pressure ifthe volume is 50 in3 and the temperature remain fixed at 80 K.

36. The length, width, and height of a rectangular box are ℓ = 5, w = 2, and h = 3,respectively.

(a) Find the instantaneous rate of change of volume of the box with respect tothe length if w and h are held constant.

(b) Find the instantaneous rate of change of volume of the box with respect tothe width if ℓ and h are held constant.

(c) Find the instantaneous rate of change of volume of the box with respect tothe height if ℓ and w are held constant.

37. The area A of a triangle is given by A = 12ab sin θ, where a and b are the lengths

of two sides and θ is the angle between these sides. Suppose that a = 5, b = 10,and θ = π/3.


(a) Find the rate at which A changes with respect to a if b and θ are held constant.

(b) Find the rate at which A changes with respect to θ if a and b are held constant.

(c) Find the rate at which b changes with respect to a if A and θ are held constant.

38. (a) By differentiating implicitly, find the slope of the hyperboloid x2 + y2 − z2 = 1in the x-direction at the points (3, 4, 2

√6) and (3, 4,−2

√6).

(b) Check the result in part (a) by solving for z and differentiating the resultingfunctions directly.

39. (a) By differentiating implicitly, find the slope of the hyperboloid x2 + y2 − z2 = 1in the y-direction at the points (3, 4, 2

√6) and (3, 4,−2

√6).

(b) Check the result in part (a) by solving for z and differentiating the resultingfunctions directly.

40− 43 Calculate ∂z/∂x and ∂z/∂y using implicit differentiation. Leave youranswers in terms of x, y, and z.

40. (x2 + y2 + z2)3/2 = 1 41. ln(2x2 + y − z3) = x

42. x2 + z sin xyz = 0 43. exy sinh z − z2x+ 1 = 0

44− 47 Find ∂w/∂x, ∂w/∂y, and ∂w/∂z using implicit differentiation. Leaveyour answers in terms of x, y, z, and w.

44. (x2 + y2 + z2 + w2)3/2 = 4 45. ln(2x2 + y − z3 + 3w) = z

46. w2 + w sin xyz = 1 47. exy sinhw − z2w + 1 = 0

48. Let z =√x cos y. Find

(a) ∂2z/∂x2 (b) ∂2z/∂y2 (c) ∂2z/∂x∂y (d) ∂2z/∂y∂x

49. Let f(x, y) = 4x2 − 2y + 7x4y5. Find

(a) fxx (b) fyy (c) fxy (d) fyx

50. Given f(x, y) = x3y5 − 2x2y + x, find

(a) fxxy (b) fyxy (c) fyyy

51. Given z = (2x− y)5, find

(a)∂3z

∂y∂x∂y(b)

∂3z

∂x2∂y(c)

∂4z

∂x2∂y2

52. Given f(x, y) = y3e−5x, find

(a) fxxy(0, 1) (b) fxxx(0, 1) (c) fyyxx(0, 1)

53. Given w = ey cosx, find

(a)∂3w

∂y2∂x

∣

∣

∣

∣

(π/4,0)

(b)∂3w

∂x2∂y

∣

∣

∣

∣

(π/4,0)


54. Let f(x, y, z) = x3y5z7 + xy2 + y3z. Find

(a) fxy (b) fyz (c) fxz (d) fzz

(e) fzyy (f) fxxy (g) fzyx (h) fxxyz

55. Let w = (4x− 3y + 2z)5, find

(a)∂2w

∂x∂z(b)

∂3w

∂x∂y∂z(c)

∂4w

∂z2∂y∂x

56− 67 Compute the differential dz or dw of the specified function.

56. z = 7x− 2y 57. z = exy

58. z = x3y2 59. z = 5x2y5 − 2x+ 4y + 7

60. z = tan−1 xy 61. z = sec2(x− 3y)

62. w = 8x− 3y + 4z 63. w = exyz

64. w = x3y2z 65. 4x2y3z7 − 3xy + z − 5

66. w = tan−1(xyz) 67. w =√x+

√y +

√z

68− 75 (a) Find the local linear approximation L to the specified function fat the designated point P . (b) Use it to estimate f at the specified point Q.

68. f(x, y) =1

√

x2 + y2; P (4, 3), Q(3.92, 3.01)

69. f(x, y) = x0.5y0.3; P (1, 1), Q(1.05, 0.97)

70. f(x, y) = x sin y; P (0, 0), Q(0.003, 0.004)

71. f(x, y) = ln xy; P (1, 2), Q(1.01, 2.02)

72. f(x, y, z) = xyz; P (1, 2, 3), Q(1.001, 2.002, 3.003)

73. f(x, y, z) =x+ y

y + z; P (−1, 1, 1), Q(−0.99, 0.99, 1.01)

74. f(x, y, z) = xeyz ; P (1,−1,−1), Q(0.99,−1.01,−0.99)

75. f(x, y, z) = ln(x+ yz); P (2, 1,−1), Q(2.02, 0.97,−1.01)


1. (a) 9x2y2 (b) 6x3y (c) 9y2 (d) 9x2 (e) 6y (f) 6x3 (g) 36 (h) 12

3. (a) 38

(b) 14

5. (a) −4 cos 7 (b) 2 cos 7 7. 8xy3ex2y3 , 12x2y2ex

2y3

9. x3/(y3/5 + x) + 3x2 ln(1 + xy−3/5),−35x4/(y8/5 + xy) 11. −y(x2 − y2)

(x2 + y2)2,x(x2 − y2)

(x2 + y2)2


13. (3/2)x2y(5x2 − 7)(3x5y − 7x3y)−1/2, (1/2)x3(3x2 − 7)(3x5y − 7x3y)−1/2

15.y−1/2

y2 + x2,− xy−3/2

y2 + x2− 3

2y−5/2 tan−1

(

x

y

)

17. −43y2 sec2 x(y2 tan x)−7/3,−8

3y tan x(y2 tan x)−7/3 18. −6,−21 20. 1/

√17, 8/

√17

22. (a) 2xy4z3 + y (b) 4x2y3z3 (c) 3x2y4z2 + 2z (d) 2y4z3 + y (e) 32z3 + 1 (f)438

24. 2z/x, z/y, ln(x2y cos z)− z tan z

26. −y2z3/(1 + x2y4z6),−2xy3/(1 + x2y4z6),−3xy2z2/(1 + x2y4z6)

27. yzez cos(xz), ez sin(xz), yez(sin(xz) + x cos(xz))

29. x/√

x2 + y2 + z2, y/√

x2 + y2 + z2, z/√

x2 + y2 + z2

31. (a) e (b) 2e (c) e 33. (a) ∂V/∂r = 2πrh (b) ∂V/∂h = πr2 (c) 48π (d) 64π

35. (a)1

5

lb

in2 ·K (b) −25

8

in5

lb36. (a)

∂V

∂t= 6 (b)

∂V

∂w= 15 (c)

∂V

∂h= 10

38. (a) ±√6/4 40. −x/z,−y/z

42. − 2x+ yz2 cos(xyz)

xyz cos(xyz) + sin(xyz);− xz2 cos(xyz)

xyz cos(xyz) + sin(xyz)

44. −x/w,−y/w,−z/w 46. − yzw cos(xyz)

2w + sin(xyz),− xzw cos(xyz)

2w + sin(xyz),− xyw cos(xyz)

2w + sin(xyz)

48. (a) − cos y

4√x3

(b) −√x cos y (c) − 1

2√xsin y (d) − 1

2√xsin y

52. (a) −30 (b) −125 (c) 150

54. (a) 15x2y4z7 + 2y (b) 35x3y4z6 + 3y2 (c) 21x2y5z6 (d) 42x3y5z5

(e) 140x3y3z6 + 6y (f) 30xy4z7 (g) 105x2y4z6 (h) 210xy4z6

56. dz = 7 dx−2 dy 58. dz = 3x2y2 dx+2x3y dy 60. dz =y

1 + x2y2dx+

x

1 + x2y2dy

62. dw = 8 dx− 3 dy + 4 dz 64. dw = 3x2y2z dx+ 2x3yz dy + x3y2 dz

66. dw =yz

1 + x2y2z2dx+

xz

1 + x2y2z2dy +

xy

1 + x2y2z2dz

68. (a) L =1

5− 4

125(x− 4)− 3

125(y − 3) 70. (a) L = 0

72. (a) L = 6+6(x−1)+3(y−2)+2(z−3) 74. (a) L = e+e(x−1)−e(y+1)−e(z+1)


5.4 The Chain Rule

The Chain Rule for Derivatives

If x = x(t) and y = y(t) are differentiable at t, and if z = f(x, y) is differentiableat the point (x, y) = (x(t), y(t)), then z = f(x(t), y(t)) is differentiable at t and

dz

dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt(5.9)

where the ordinary derivatives are evaluated at t and the partial derivatives areevaluated at (x, y).

Theorem 5.7 (Chain Rule for Derivatives)

Formula (5.9) can be represented schematically by a “tree diagram” that is con-structed as follows.

z

t t

x y

∂z∂x

∂z∂y

dxdt

dydt

dz

dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt

Suppose thatz = x2ey, x(t) = t2 − 1, y(t) = sin t

Use the chain rule to find dz/dt.

Example 5.29

Solution


A simple electrical circuit consists of a resistor R and electromotive force V .At a centain instant V is 80 volts and is increasing at a rate of 5 volts/min,while R is 40 ohms and is decreasing at a rate of 2 ohms/min. Use Ohm’s law,I = V/R, and a chain rule to find the rate at which the current I (in amperes)is changing.

Example 5.30

Solution

If each of the functions x = x(t), y = y(t), and z = z(t) is differentiable at t,and if w = f(x, y, z) is differentiable at the point (x, y, z) = (x(t), y(t), z(t)),then w = f(x(t), y(t), z(t)) is differentiable at t and

dw

dt=

∂w

∂x

dx

dt+

∂w

∂y

dy

dt+

∂w

∂z

dz

dt(5.10)

where the ordinary derivatives are evaluated at t and the partial derivatives areevaluated at (x, y, z).

Theorem 5.8 (Chain Rule for Derivatives)

The following figure shows tree diagrams for the formulas in Theorem 5.8.

w

x y z

∂w∂x

∂w∂y

∂w∂z

t t t

dxdt

dydt

dzdt

dw

dt=

∂w

∂x

dx

dt+

∂w

∂y

dy

dt+

∂w

∂z

dz

dt


Use the chain rule to find dw/dt if

w = x2 + yz, x = 3t2 + 1, y = 2t− 4, z = t3.

Example 5.31

Solution

Suppose that

w =√

x2 + y2 + z2, x = cos θ, y = sin θ, z = tan θ

Use the chain rule to find dw/dθ.

Example 5.32

Solution


The Chain Rule for Partial Derivatives

If x = x(u, v) and y = y(u, v) have first-order derivatives at the point (u, v),and if z = f(x, y) is differentiable at the point (x, y) = (x(u, v), y(u, v)), thenz = f(x(u, v), y(u, v)) has first-order partial derivatives at the point (u, v) givenby

∂z

∂u=

∂z

∂x

∂x

∂u+

∂z

∂y

∂y

∂uand

∂z

∂v=

∂z

∂x

∂x

∂v+

∂z

∂y

∂y

∂v

Theorem 5.9 (Chain Rule for Partial Derivatives)

The following figure shows tree diagrams for the formulas in Theorem 5.9.

z

u v vu

x y

∂z∂x

∂z∂y

∂x∂u

∂x∂v

∂y∂u

∂y∂v

∂z

∂u=

∂z

∂x

∂x

∂u+

∂z

∂y

∂y

∂u

z

u v vu

x y

∂z∂x

∂z∂y

∂x∂u

∂x∂v

∂y∂u

∂y∂v

∂z

∂v=

∂z

∂x

∂x

∂v+

∂z

∂y

∂y

∂v

Given that z = exy, x(u, v) = 3u sin v, and y(u, v) = 4v2u, find ∂z/∂u and∂z/∂v using the chain rule.

Example 5.33

Solution


If each function x = x(u, v), y = y(u, v), and z = z(u, v) have first-orderderivatives at the point (u, v), and if the function w = f(x, y, z) is differ-entiable at the point (x, y, z) = (x(u, v), y(u, v), z(u, v)), then the functionw = f(x(u, v), y(u, v), z(u, v)) has first-order partial derivatives at the point(u, v) given by

∂w

∂u=

∂w

∂x

∂x

∂u+

∂w

∂y

∂y

∂u+

∂w

∂z

∂z

∂u

and∂w

∂v=

∂w

∂x

∂x

∂v+

∂w

∂y

∂y

∂v+

∂w

∂z

∂z

∂v

Theorem 5.10 (Chain Rule for Partial Derivatives)

w

x y z

∂w∂x

∂w∂y

∂w∂z

u v u v vu

∂x∂u

∂x∂v

∂y∂u

∂y∂v

∂z∂u

∂z∂v

∂w

∂u=

∂w

∂x

∂x

∂u+

∂w

∂y

∂y

∂u+

∂w

∂z

∂z

∂u

∂w

∂v=

∂w

∂x

∂x

∂v+

∂w

∂y

∂y

∂v+

∂w

∂z

∂z

∂v

Suppose that w = x+2y+z2, x =u

v, y = u2+ln v, and z = 2u. Use appropriate

forms of the chain rule to find ∂w/∂u and ∂w/∂v.

Example 5.34

Solution


Other Versions of the Chain Rule

Although we will not prove it, the chain rule extends to functions w = f(v1, v2, . . . , vn)of n variables. For example, if each vi is a function of t, i = 1, 2, . . . , n, the relevantformula is

dw

dt=

∂w

∂v1

dv1dt

+∂w

∂v2

dv2dt

+ · · ·+ ∂w

∂vn

dvndt

(5.11)

There are infinitely many variations of the chain rule, depending on the numberof variables and the choice of independent and dependent variables. A good workingprocedure is to use tree diagrams to derive new versions of the chain rule as needed.

Suppose that w = x2 + y2 + z2 and

x = ρ sin φ cos θ, y = ρ sinφ sin θ, z = ρ cos φ

Use appropriate forms of the chain rule to find ∂w/∂ρ and ∂w/∂θ.

Example 5.35

Solution


Suppose that w = xy + yz, y = sin x, and z = ex. Use appropriate form of thechain rule to find dw/dx.

Example 5.36

Solution

Implicit Differentiation

Consider the special case where z = f(x, y) is a function of x and y and y is a differen-tiable function of x. Equation (5.9) then becomes

dz

dx=

∂f

∂x

dx

dx+

∂f

∂y

dy

dx=

∂f

∂x+

∂f

∂y

dy

dx(5.12)

This result can be used to find derivatives of functions that are defined implicitly.

If the equation f(x, y) = c defines y implicitly as a differentiable function of x,and if ∂f/∂y 6= 0, then

dy

dx= −∂f/∂x

∂f/∂y(5.13)

Theorem 5.11

Given that x3+y2x−3 = 0, find dy/dx using (5.13), and check the result usingimplicit differentiation.

Example 5.37

Solution


If the equation f(x, y, z) = c defines z implicitly as a differentiable function ofx and y, and if ∂f/∂z 6= 0, then

∂z

∂x= −∂f/∂x

∂f/∂zand

∂z

∂y= −∂f/∂y

∂f/∂z

Theorem 5.12

Consider the sphere x2 + y2 + z2 = 1. Find ∂z/∂x and ∂z/∂y at the point(

23, 13, 23

)

.

Example 5.38

Solution

Exercise 5.4

1− 6 Use an appropriate form of the chain rule to find dz/dt.

1. z = 3x2y3; x = tt, y = t2 2. z = ln(2x2 + y); x =√t, y = t2/3

3. z = 3 cosx− sin xy; x = 1/t, y = 3t 4. z =√

1 + x− 2xy4; x = ln t, y = t

5. z = e1−xy; x = t1/3, y = t3

6− 9 Use an appropriate form of the chain rule to find dw/dt.

6. w = 5x2y3z4; x = t2, y = t3, z = t5

7. w = ln(3x2 − 2y + 4z3); x = t1/2, y = t2/3, z = t−2

8. w = 5 cosxy − sin xz; x = 1/t, y = t, z = t3

9. w =√

1 + x− 2yz4x; x = ln t, y = t, z = 4t


10− 15 Use an appropriate forms of the chain rule to find ∂z/∂u and ∂z/∂v.

10. z = 8x2y − 2x+ 3y; x = uv, y = u− v

11. z = x2 − y tan x; x = u/v, y = u2v2

12. z = x/y; x = 2 cosu, y = 3 sin v

13. z = 3x− 2y; x = y + v ln u, y = u2 − v ln v

14. z = ex2y; x =

√uv, y = 1/v

15. z = cosx sin y; x = u− v, y = u2 + v2

16− 23 Use an appropriate forms of the chain rule to find the derivatives.

16. Let T = x2y − xy3 + 2; x = r cos θ, y = r sin θ. Find ∂T/∂r and ∂T/∂θ.

17. Let R = e2s−t2 ; s = 3φ, t = φ1/2. Find dR/dφ.

18. Let t = u/v; u = x2 − y2, v = 4xy3. Find ∂t/∂x and ∂t/∂y.

19. Let w = rs/(r2 + s2); r = uv, s = u− 2v. Find ∂w/∂u and ∂w/∂v.

20. Let z = ln(x2 + 1), where x = r cos θ. Find ∂z/∂r and ∂z/∂θ.

21. Let u = rs2 ln t; r = x2, s = 4y + 1, t = xy3. Find ∂u/∂x and ∂u/∂y.

22. Let w = 4x2 + 4y2 + z2, x = ρ sinφ cos θ, y = ρ sin φ sin θ, z = ρ cos φ. Find∂w/∂ρ, ∂w/∂φ, and ∂w/∂θ.

23. Let w = 3xy2z3, y = 3x2 + 2, z =√x− 1. Find dw/dx.

24. Use a chain rule to find the value ofdw

ds

∣

∣

∣

∣

s=1/4

if w = r2 − r tan θ; r =√s, θ = πs.

25. Use a chain rule to find the value of

∂f

∂u

∣

∣

∣

∣

u=1,v=−2

and∂f

∂v

∣

∣

∣

∣

u=1,v=−2

if f(x, y) = x2y2 − x+ 2y; x =√u, y = uv3.

26. Use a chain rule to find the value of

∂z

∂r

∣

∣

∣

∣

r=2,θ=π/6

and∂z

∂θ

∣

∣

∣

∣

r=2,θ=π/6

if z = xyex/y; x = r cos θ, y = r sin θ.

27. Use a chain rule to finddz

dt

∣

∣

∣

∣

t=3

if z = x2y; x = t2, y = t+ 7.


28− 31 Use Theorem14.4 to find dy/dx and check your result usingimplicit differentiation.

28. x2y3 + cos y = 0 29. x3 − 3xy2 + y3 = 5

30. exy + yey = 1 31. x−√xy + 3y = 4

32. Two straight roads intersect at right angles. Car A, moving on one of the roads,approaches the intersection at 25 mi/h and car B, moving on the other road,approaches the intersection at 30 mi/h. At what rate is the distance between thecars changing when A is 0.3 mile from the intersection and B is 0.4 mile from theintersection?

33. Use the ideal gas law P = kT/V with V in cubic inches (in3), T in kelvins (K),and k = 10 in · lb/K to find the rate at which the temperature of a gas is changingwhen the volume is 200 in3 and increasing at the rate of 4 in3/s, while the pressureis 5 lb/in2 and decreasing at the rate of 1 lb/in2/s.

34. Two sides of a triangle have lengths a = 4 cm and b = 3 cm but are increasing atthe rate of 1 cm/s. If the area of the triangle remains constant, at what rate is theangle θ between a and b changing when θ = π/6?

35. Two sides of a triangle have lengths a = 5 cm and b = 10 cm, and the includedangle is θ = π/3. If a is increasing at a rate of 2 cm/s, b is increasing at a rateof 1 cm/s, and θ remains constant, at what rate is the third side changing? Is itincreasing or decreasing? [Hint: Use the law of cosines.]

36. The length, width, and height of a rectangular box are increasing at rates of 1 in/s,2 in/s, and 3 in/s, respectively.

(a) At what rate is the volume increasing when the length is 2 in, the width is 3in, and the height is 6 in?

(b) At what rate is the length of the diagonal increasing at that instant?

37. Consider the box in Exercise 36. At what rate is the surface area of the boxincreasing at the given instant?


1. 42t13 3. 3t−2 sin(1/t) 5. −10

3t7/3e1−t10/3 6.

dw

dt= 165t32 8. −2t cos t2

10. 24u2v2 − 16uv3 − 2v + 3, 16u3v − 24u2v2 − 2u− 3 12. −2 sin u

3 sin v, −2 cosu cos v

3 sin2 v

14. eu, 0

16. 3r2 sin θ cos2 θ−4r3 sin3 θ cos θ, −2r3 sin2 θ cos θ+r4 sin4 θ+r3 cos3 θ−3r4 sin2 θ cos2 θ

18.x2 + y2

4x2y3,y2 − 3x2

4xy420.

∂z

∂r=

2r cos2 θ

r2 cos2 θ + 1,∂z

∂θ=

−2r2 cos θ sin θ

r2 cos2 θ + 1


22.dw

dρ= 2ρ

(

4 sin2 φ+ cos2 φ)

,dw

dφ= 6ρ2 sinφ cosφ,

dw

dθ= 0 24. −π

26.√3e

√3,

(

2− 4√3)

e√3

5.5 Directional Derivatives and Gradients

Directional Derivatives

If f(x, y) is a function of x and y, and if ~u = 〈u1, u2〉 is a unit vector, then thedirectional derivative of f in the direction of ~u at (a, b) is denoted byD~uf(a, b) and is defined by

D~uf(a, b) =d

ds

[

f(a+ su1, b+ su2)]

s=0(5.14)

provided this derivative exists.

Definition 5.9

Geometrically, D~uf(a, b) can be interpreted as the slope of the surface z = f(x, y)in the direction of ~u at the point (a, b, f(a, b)).

Usually the value of D~uf(a, b) will depend on both the point (a, b) and the direction~u. Thus, at a fixed point the slope of the surface may vary with the direction.

The definition of a directional derivative for a function f(x, y, z) of three variables issimilar to Definition 5.9.

If ~u = 〈u1, u2, u3〉 is a unit vector, and if f(x, y, z) is a function of x, y, and z,then the directional derivative of f in the direction of ~u at (a, b, c) isdenoted by D~uf(a, b, c) and is defined by

D~uf(a, b, c) =d

ds[f(a+ su1, b+ su2, c+ su3)]s=0 (5.15)

provided this derivative exists.

Definition 5.10

For a function that is differentiable at a point, directional derivatives exist in everydirection from the point and can be computed directly in terms of the first-order partialderivatives of the function.


(a) If f(x, y) is differentiable at (a, b) and if ~u = 〈u1, u2〉 is a unit vector, thenthe directional derivative D~uf(a, b) exists and is given by

D~uf(a, b) = fx(a, b)u1 + fy(a, b)u2 (5.16)

(b) If f(x, y, z) is differentiable at (a, b, c) and if ~u = 〈u1, u2, u3〉 is a unitvector, then the directional derivative D~uf(a, b, c) exists and is given by

D~uf(a, b, c) = fx(a, b, c)u1 + fy(a, b, c)u2 + fz(a, b, c)u3 (5.17)

Theorem 5.13

For f(x, y) = x2y − 4y3, compute D~uf(2, 1) where ~u is a unit vector in thedirection from (2, 1) to (4, 0).

Example 5.39

Solution


Recall that a unit vector ~u in the xy-plane can be expressed as

~u = cos θ~i+ sin θ~j

where θ is the angle from the positive x-axis to ~u. Thus, Formula (5.16) can also beexpressed as

D~uf(a, b) = fx(a, b) cos θ + fy(a, b) sin θ (5.18)

Find the directional derivative of f(x, y) = exy at (−2, 0) in the direction of theunit vector that makes an angle of π/3 with the positive x-axis.

Example 5.40

Solution

Find the directional derivative of f(x, y, z) = x2y − yz3 + z at (1,−2, 0) in the

direction of the vector ~a = 2~i+~j − 2~k.

Example 5.41

Solution


The Gradient

(a) If f is a function of x and y, then the gradient of f is defined by

∇f(x, y) = fx(x, y)~i+ fy(x, y)~j (5.19)

(b) If f is a function of x, y, and z, then the gradient of f is defined by

∇f(x, y, z) = fx(x, y, z)~i+ fy(x, y, z)~j + fk(x, y, z)~k (5.20)

Definition 5.11

Formulas (5.16) and (5.17) can now be written as

D~uf(a, b) = ∇f(a, b) · ~u (5.21)

andD~uf(a, b, c) = ∇f(a, b, c) · ~u (5.22)

respectively.

Formula (5.21) can be interpreted to mean that the slope of the surface z = f(x, y)at the point (a, b) in the direction of ~u is the dot product of the gradient with ~u.

Properties of the Gradient

Let f be a function of either two variables or three variables, and let P denotethe point P (a, b) or P (a, b, c), respectively. Assume that f is differentiable atP .

(a) If ∇f = ~0 at P , then all directional derivatives of f at P are zero.

(b) If ∇f 6= ~0 at P , then among all possible directional derivatives of f atP , the derivative in the direction of ∇f at P has the largest value. Thevalue of this largest directional derivative is ‖∇f‖ at P .

(c) If ∇f 6= ~0 at P , then among all possible directional derivatives of f atP , the derivative in the direction opposite to that of ∇f at P has thesmallest value. The value of this smallest directional derivative is −‖∇f‖at P .

Theorem 5.14


Let f(x, y) = x2ey. Find the maximum value of a directional derivative at(−2, 0), and find the unit vector in the direction in which the maximum valueoccurs.

Example 5.42

Solution

Exercise 5.5

1− 8 Find D~uf at P .

1. f(x, y) = (1 + xy)3/2 ; P (3, 1) ; ~u =1√2~i+

1√2~j

2. f(x, y) = exy ; P (4, 0) ; ~u = −35~i+ 4

5~j

3. f(x, y) = ln(1 + x2 + y) ; P (0, 0) ; ~u = − 1√10

~i− 3√10

~j

4. f(x, y) =cx+ dy

x− y; P (3, 4) ; ~u = 4

5~i+ 3

5~j

5. f(x, y, z) = 4x5y2z3 ; P (2,−1, 1) ; ~u = 13~i+ 2

3~j − 2

3~k

6. f(x, y, z) = yexz + z2 ; P (0, 2, 3) ; ~u = 27~i− 3

7~j + 6

7~k

7. f(x, y, z) = ln(x2 + 2y2 + 3z2) ; P (−1, 2, 4) ; ~u = − 313~i− 4

13~j − 12

13~k

8. f(x, y, z) = sin xyz ; P(

12, 13, π

)

; ~u =1√3~i− 1√

3~j +

1√3~k

9− 18 Find the directional derivative of f at P in the direction of ~a.

9. f(x, y) = 4x3y2 ; P (2, 1) ; ~a = 4~i− 3~j


10. f(x, y) = x2 − 3xy + 4y3 ; P (−2, 0) ; ~a =~i+ 2~j

11. f(x, y) = y2 ln x ; P (1, 4) ; ~a = 3~i+ 3~j

12. f(x, y) = ex cos y ; P (0, π/4) ; ~a = 5~i− 2~j

13. f(x, y) = tan−1(y/x) ; P (−2, 2) ; ~a = −~i−~j

14. f(x, y) = xey − yex ; P (0, 0) ; ~a = 5~i− 2~j

15. f(x, y, z) = x3z − yx2 + z2 ; P (2,−1, 1) ; ~a = 3~i−~j + 2~k

16. f(x, y, z) = y −√x2 + z2 ; P (−3, 1, 4) ; ~a = 2~i− 2~j − ~k

17. f(x, y, z) =z − x

z + y; P (1, 0,−3) ; ~a = −6~i+ 3~j − 2~k

18. f(x, y, z) = ex+y+3z ; P (−2, 2,−1) ; ~a = 20~i− 4~j + 5~k

19− 21 Find the directional derivative of f at P in the direction of a vectormaking the counterclockwise angle θ with the positive x-axis.

19. f(x, y) =√xy ; P (1, 4) ; θ = π/3

20. f(x, y) =x− y

x+ y; P (−1,−2) ; θ = π/2

21. f(x, y) = tan(2x+ y) ; P (π/6, π/3) ; θ = 7π/4

22. Find the directional derivative of f(x, y) =x

x+ yat P (1, 0) in the direction of

Q(−1,−1).

23. Find the directional derivative of f(x, y) = e−x sec y at P (0, π/4) in the directionof the origin.

24. Find the directional derivative of f(x, y) =√xy ey at P (1, 1) in the direction of

the negative y-axis.

25. Letf(x, y) =

y

x+ y

Find a unit vector ~u for which D~uf(2, 3) = 0.

26. Find the directional derivative of

f(x, y, z) =y

x+ z

at P (2, 1,−1) in the direction form P to Q(−1, 2, 0).

27. Find the directional derivative of the function

f(x, y, z) = x3y2z5 − 2xz + yz + 3x

at P (−1,−2, 1) in the direction of the negative z-axis.


28− 31 Find ∇z or ∇w.

28. z = sin(7y2 − 7xy) 29. z =6x+ 7y

6x− 7y

30. w = −x9 − y3 + z12 31. w =√

x2 + y2 + z2

32− 37 Find the gradient of f at the indicate point.

32. f(x, y) = 5x2 + y4 ; (4, 2) 33. f(x, y) = 5 sin x2 + cos 3y ; (√π/2, 0)

34. f(x, y) = (x2 + xy)3 ; (−1,−1) 35. f(x, y) = (x2 + y2)−1/2 ; (3, 4)

36. f(x, y, z) = y ln(x+ y + z) ; (−3, 4, 0) 37. f(x, y, z) = y2z tan3 x ; (π/3,−3, 1)


1. 6√2 3. −3/

√10 5. −320 7. −314/741 9. 0 11. −8

√2 13.

√2/4

17. −8/63 19. 1/2 +√3/8 21. 2

√2 22. 1/

√5 24. −3

2e 26. 3/

√11

28. cos(7y2 − 7xy)(

− 7~i+ (14y − 7x)~j)

29.

( −84y

(6x− 7y)2

)

~i+

(

84x

(6x− 7y)2

)

~j

30. −9x8~i− 3y2~j + 12z11 ~k

31. ∇w =x

x2 + y2 + z2~i+

y

x2 + y2 + z2~j +

z

x2 + y2 + z2~k

32. 40~i+ 32~j 34. −36~i− 12~j 36. 4(~i+~j + ~k)

5.6 Maxima and Minima of Functions of Two Vari-

ables

In the previous course we learned how to find maximum and minimum values of afunction of one variable. In this section we will develop similar techniques for functionsof two variables.

Extrema

If we imagine the graph of a function f of two variables to be a mountain range (Figurebelow), then the mountaintops, which are the high points in their immediate vicinity,are called local maxima of f , and the valley bottoms, which are the low points in theirimmediate vicinity, are called local minima of f .


A function f of two variables is said to have a local maximum at a point (a, b)if there is a disk centered at (a, b) such that f(a, b) ≥ f(x, y) for all points (x, y)that lie inside the disk, and f is said to have an absolute maximum at (a, b)if f(a, b) ≥ f(x, y) for all points (x, y) in the domain of f .

Definition 5.12

A function f of two variables is said to have a local minimum at a point (a, b)if there is a disk centered at (a, b) such that f(a, b) ≤ f(x, y) for all points (x, y)that lie inside the disk, and f is said to have an absolute minimum at (a, b)if f(a, b) ≤ f(x, y) for all points (x, y) in the domain of f .

Definition 5.13

x

y

zf(a, b)

f(x, y)

(a, b) (x, y)

local maximum f(a, b)

x

y

z

f(a, b)

f(x, y)

(a, b)(x, y)

local minimum f(a, b)

If f has a local maximum or a local minimum at (a, b), then we say that f has alocal extremum at (a, b), and if f has an absolute maximum or absolute minimum at(a, b), then we say that f has an absolute extremum at (a, b).


Finding Local Extrema

If f has a relative extremum at a point (a, b), and if the first order partialderivatives of f exist at this point, then

fx(a, b) = 0 and fy(a, b) = 0.

Theorem 5.15

Recall that the critical points of a function f of one variable are those values of x inthe domain of f at which f ′(x) = 0 or f is not differentiable. The following definition isthe analog for functions of two variables.

A point (a, b) in the domain of a function f(x, y) is called a critical point ofthe function if fx(a, b) = 0 and fy(a, b) = 0 or if one or both partial derivativesdo not exist at (a, b).

Definition 5.14

Find all critical points of f(x, y) = xe−x2/2−y3/3+y.

Example 5.43

Solution


It follows from Definition 5.14 and Theorem 5.15 that local extrema occur at criticalpoints, just as for a function of one variable. However, recall that for a function of onevariable a local extremum need not occur at every critical point. Similarly, a functionof two variables need not have a local extremum at every critical point. For example,consider the function

f(x, y) = y2 − x2

This function, whose graph is the hyperbolic paraboloid shown in Figure below, has acritical point at (0, 0), since

fx(x, y) = −2x and fy(x, y) = 2y

from which it follows that

fx(0, 0) = 0 and fy(0, 0) = 0.

x

z

yb(0, 0)

However, the function f has neither a local maximum nor a local minimum at (0, 0).The point (0, 0) is called a saddle point of f .

A differentiable function f(x, y) has a saddle point at a critical point (a, b) ifthere is an open disk R containing (a, b) such that f(x, y) > f(a, b) for somepoints (x, y) in R and f(x, y) < f(a, b) for other points.

Definition 5.15

The Second Partials Test

For functions of one variable the second derivative test was used to determine the behav-ior of a function at a critical point. The following theorem is the analog of that theoremfor functions of two variables.


Let f be a function of two variables with continuous second-order partial deriva-tives in some disk centered at a critical point (a, b), and let

D(a, b) = fxx(a, b)fyy(a, b)−[

fxy(a, b)]2.

(a) If D(a, b) > 0 and fxx(a, b) > 0, then f has a local minimum at (a, b).

(b) If D(a, b) > 0 and fxx(a, b) < 0, then f has a local maximum at (a, b).

(c) If D(a, b) < 0, then f has a saddle point at (a, b).

(b) If D(a, b) = 0, then no conclusion can be drawn.

Theorem 5.16 (The Second Partials Test)

The expression fxxfyy −[

fxy]2

is called the discriminant or Hessian of f . It issome times easier to remember it in determinant form,

D = fxxfyy −[

fxy]2

=

∣

∣

∣

∣

fxx fxyfxy fyy

∣

∣

∣

∣

.

Locate all local extrema and saddle points of

f(x, y) = xy − x2 − y2 − 2x− 2y + 4.

Example 5.44

Solution


Locate all local extrema and saddle points of

f(x, y) = x3 + 2xy − 2y2 − 10x.

Example 5.45

Solution


Exercise 5.6

1− 12 Locate all local extrema and saddle points of the following functions.

1. f(x, y) = y2 + xy + 3y + 2x+ 3

2. f(x, y) = x2 + xy + y2 − 3x

3. f(x, y) = x2 + 4x2 − x+ 2y

4. f(x, y) = x2 + 2y2

5. f(x, y) = x3 − 3xy + y3

6. f(x, y) = x2 + y2 +2

xy

7. f(x, y) = x2 + y − ey

8. f(x, y) =4y + x2y2 + 8x

xy

9. f(x, y) = ex sin y

10. f(x, y) = e−x2−y2

11. f(x, y) = e−(x2+y2+2x)

12. f(x, y) = xe−x2−y2


1. saddle point: (1,−2) 2. local minimum: (2,−1) 3. local minimum:(

12,−1

4

)

4. local minimum: (0, 0) 5. saddle point: (1,−2), local minimum: (1, 1)

6. local minimum: (−1,−1), (1, 1) 7. saddle point: (0, 0)

8. local minimum:(

3√2, 2 3

√2)

9. No critical points 10. local maximum: (0, 0)

11. local maximum: (−1, 0) 12. saddle point:(

±√2, 0

)

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Chapter 5 Partial Derivatives