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 independent variables occur. For instance, the ideal gas law P = ρRT states that the pressure P is a function of both its density ρ and its temperature T . (The gas constant R is a material property and not a variable). In this section, we expand our concept of function to include functions that depend on 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 for functions 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 dependent variable 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 (x 1 ,x 2 ,...,x n ) expresses u as a function of x 1 ,x 2 ,...,x n . 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 unique real number f (x, y ) to each point (x, y ) in some set D in the xy -plane. Definition 5.1 129
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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
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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
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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
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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
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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).
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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.
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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
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Evaluate lim(x,y)→(0,0)
x3 − y3
x− y.
Example 5.6
Solution
Evaluate lim(x,y)→(0,0)
x2 − xy√x−√
y.
Example 5.7
Solution
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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)
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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
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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
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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
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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
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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
Answers to Exercise 5.2
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.
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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
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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)
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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).
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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).
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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
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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
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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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 151
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 152
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 153
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).
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 154
Similarly, for a function w = f(x, y, z) of three variables we have the total differ-
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 155
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).
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.
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 157
(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.
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 158
(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)
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 159
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)
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)
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 161
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 162
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 163
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 164
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 165
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 166
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 167
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 168
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 169
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.
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 170
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?
Answers to Exercise 5.4
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 171
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.
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 172
(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
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 173
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 174
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 175
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.
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)
Answers to Exercise 5.5
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 .
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 178
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).
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 179
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 180
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.
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 181
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
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 182
Locate all local extrema and saddle points of
f(x, y) = x3 + 2xy − 2y2 − 10x.
Example 5.45
Solution
MA112: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 183
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
Answers to Exercise 5.6
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)