math 355: calculus of several variables · 1|partial di erentiation 1.1partial derivatives de...

MATH 355: Calculus of SeveralVariables

Dr. Joseph K. Ansong

ii c©Dr Joseph K. Ansong

Contents

1 Partial Di�erentiation 1

1.1 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Interpretation of Partial Derivatives . . . . . . . . . . . . . . . . . . 3

1.1.2 Higher Order Partial Derivatives . . . . . . . . . . . . . . . . . . . 8

iii

CONTENTS

iv c©Dr Joseph K. Ansong

1 | Partial Di�erentiation

1.1 Partial Derivatives

De�nition 1. If f is a function of two variables, its partial derivatives are denotedby fx and fy and de�ned by

fx(x, y) = limh→0

f(x+ h, y)− f(x, y)h

fy(x, y) = limh→0

f(x, y + h)− f(x, y)h

To obtain the partial derivatives at a point, say (a, b), we just substitude the point intothe formulas above.

Remark 1. Other notations for partial derivatives are given below, where z = f(x, y),

fx(x, y) = fx =∂f

∂x=∂z

∂x= f1 = D1f = Dxf

fx(x, y) = fy =∂f

∂y=∂z

∂y= f2 = D2f = Dyf

PowerPoint To �nd the partial derivative with respect to one variable, we simply regard the othervariables as constants. For instance, to �nd fx, we regard y as a constant, and to �nd fy,we regard x as a constant.

Example 1. Find fx and fy of the following functions.

(a) f(x, y) = x3 + 2xy − y3

(b) f(x, y) = y√x− ln (x2y)− x

y

(c) f(x, y) = y + cos(xy)− exy

(d) f(x, y) = y2x+ sin

(5x

2 + y

)Solution 1. (a) f(x, y) = x3 + 2xy − y3

fx = 3x2 + 2y − 0 = 3x2 + 2y

fy = 0 + 2x− 3y2 = 2x− 3y2

1

CHAPTER 1. PARTIAL DIFFERENTIATION

(b) f(x, y) = y√x− ln (x2y)− x

y

fx =1

2yx−1/2 − 2xy

x2y− 1

y=

y

2√x− 2

x− 1

y

fy =√x− x2

x2y+x

y2=√x− 1

y+x

y2

(c) f(x, y) = y + cos(xy)− exy

fx = 0− y sin(xy)− yexy = −y sin(xy)− yexy

fy = 1− x sin(xy)− xexy

(d) f(x, y) = y2x+ sin

(5x

2 + y

)fx = y2 +

(5

2 + y

)cos

(5x

2 + y

)fy = 2yx−

(5x

(2 + y)2

)cos

(5x

2 + y

)Example 2. Evaluate fx and fy at the given point (a, b)

(a) f(x, y) = 2x4 + x sin(yπ)− xy2, (a, b) = (2, 1)

(b) f(x, y) = x√y +

xy

x+ y, (a, b) = (1, 4)

Solution 2. (a) f(x, y) = 2x4 + x sin(yπ)− xy2, (a, b) = (2, 1)

fx(x, y) = 8x3 + sin(yπ)− y2

∴ fx(2, 1) = 8(2)3 + sin(π) = 64 + 0 = 64

fy(x, y) = 0 + πx cos(yπ)− 2xy = πx cos(yπ)− 2xy

∴ fy(2, 1) = 2π cos(π)− 2(2)(1) = −2π − 4

(b) f(x, y) = x√y +

xy

x+ y, (a, b) = (1, 4)

To compute the expressionxy

x+ ywith respect x or y, we can apply either the

quotient rule or the product rule. Let's apply the product rule in the following by

writtingxy

x+ y= x · y

x+ yor

xy

x+ y= y · x

x+ y. Thus,

fx(x, y) =√y +

[y

x+ y− xy

(x+ y)2

]∴ fx(1, 4) =

√4 +

[4

5− 4

25

]= 2 +

16

25=

66

25

fy(x, y) =1

2y−1/2 +

[x

x+ y− xy

(x+ y)2

]=

1

2√y+

[x

x+ y− xy

(x+ y)2

]∴ fy(1, 4) =

1

4+

[1

5− 4

25

]=

1

4+

1

25=

29

100

2 c©Dr Joseph K. Ansong

1.1. PARTIAL DERIVATIVES

1.1.1 Interpretation of Partial Derivatives

The partial derivatives of a function of several variables represent the rates of changeof the function with respect to the variable. For instance, fx(x, y) represents the rate ofchange of f(x, y) in the x direction when y is held �xed. Similarly, fy(x, y) representsthe rate of change of f(x, y) in the y direction when x is held �xed.

Illustrative Example 1. Given that f(x, y) =2y

x2. Determine whether f is increasing

or decreasing at the point (2, 1).

Illustrative Solution 1. fx(x, y) = −4y

x3, fy(x, y) =

2

x2. At the point (2, 1), we have

fx(2, 1) = −4

8= −1

2< 0, and fy(2, 1) =

1

2> 0

Thus, fx(2, 1) < 0 implies that if we allow x to vary/change and hold y �xed, then f isdecreasing at the point (2, 1). On the other hand, fy(2, 1) > 0, means that if we allow yto vary and hold x constant, then f is increasing at the point (2, 1).

Illustrative Example 2. Consider the hyperbolic paraboloid f(x, y) = y2 − x2.Determine whether f is increasing or decreasing at the point (1, 1).

Illustrative Solution 2. fx(x, y) = −2x, fy(x, y) = 2y. At the point (1, 1), we have

fx(1, 1) = −2 < 0, and fy(1, 1) = 2 > 0

As mentioned above, because fx(1, 1) < 0, it means that if we allow x to vary and hold y�xed, then f is decreasing at the point (1, 1). And fy(1, 1) > 0 means that if we allow yto vary and hold x constant, then f is increasing at the point (1, 1). This can be seen inthe graph of the hyperbolic paraboloid in Figure 1.1

Geometrical interpretation

Consider the surface z = f(x, y) in Figure 1.2(a). The intersection of the paraboloidwith the plane y = b (in blue) yields the curve C1 (in black). Then fx(x, b) representsthe rate of change of the curve C1 along the x−axis. The curve C1 is said to be the traceof the surface z = f(x, y) in the plane y = b. The equation of the curve C1 is given byg(x) = f(x, b). So the slope of it's tangent at any position along the x−direction is givenby the partial derivative g′(x) = fx(x, b). Explicitly, the slope of the tangent to curve C1

at the point P (a, b, c) is equal to g′(a) = fx(a, b), as demonstrated in Figure 1.2(b).Similarly, the partial derivative h(y) = fy(a, y) can be interpreted as the slope of thetangent line to the trace or curve obtained by the intersection of the �xed plane x = aand the surface z = f(x, y) as displayed in Figure 1.2(c-d). A demonstration of thepartial derivatives fx(a, b) and fy(a, b) at the point P (a, b, c) is shown in Figure 1.3.

c©Dr Joseph K. Ansong 3


Figure 1.1: Graph of the hyperbolic paraboloid.

Numerical computations

Illustrative Example 3. The heat index, I, is used by meteorologists to measure thecombined e�ects of air temperature and humidity. It is the perceived air temperaturewhen the actual temperature is T and the relative humidity is H. So I is a function of Tand H such that I = f(T,H). Figure 1.4 shows the heat index for various values of Tand H. We want to

(1) estimate the rate of increase of the apparent temperature (heat index) for everydegree that the actual temperature rises, when the relative humidity is 70% andthe actual temperature is 96◦.

(2) estimate the rate of increase of the heat index for every percentage increase in therelative huminity when the actual temperature is 96◦ and the relative huminity is70%.

Illustrative Solution 3. (1) Consider the column with H = 70% in Figure 1.4. Wecan write G(T ) = f(T, 70) and gives the values of the heat index for di�erentvalues of the actual temperature when the relative humidity is 70%. The rate ofchange of I with respect to T is G′(T ) = fT (T, 70):

G′(T ) = fT (T, 70) = limh→0

G(T + h)−G(T )h

= limh→0

f(T + h, 70)− f(T, 70)h

,

where h is the di�erence in actual temperature. We can estimate the rate of



(a) (b)

(c) (d)

Figure 1.2: Interpretation of �rst partial derivatives for [(a)-(b)] fx(x, y) =∂f

∂x, and

[(c)-(d)] fy(x, y) =∂f∂y

change above when T = 96◦F as shown below using h = 2:

G′(96) = fT (96, 70) ≈f(96 + h, 70)− f(96, 70)

h=f(98, 70)− f(96, 70)

2

=133− 125

2= 4



Figure 1.3: Graph of the hyperbolic paraboloid.

Figure 1.4:

Using h = −2, we get

G′(96) = fT (96, 70) ≈f(96 + h, 70)− f(96, 70)

h=f(94, 70)− f(96, 70)

−2

=118− 125

2= 3.5

Averaging the two values gives G′(96) = fT (96, 70) ≈ (4 + 3.5)/2 = 3.75. Thisimplies that when the relative humidity is 70% and the actual temperature is96◦F , the heat index (apparent temperature) rises by about 3.75◦F for everydegree that the actual temperature increases.

(2) Now consider the row with T = 96◦F . We can write the function governing this asJ(H) = f(96, H). We can write the rate of change of I with respect to H as

J ′(H) = fH(96, H) = limh→0

J(H + h)− J(H)

h= lim

h→0

f(96, H + h)− f(96, H)

h.

When the relative humidity is H = 70%, we can estimate the rate of change of I



using h = 5 and h = −5 as

J ′(70) = fH(96, 70) ≈J(70 + 5)− J(70)

5=f(96, 75)− f(96, 70)

5=

130− 125

5= 1

J ′(70) = fH(96, 70) ≈J(70− 5)− J(70)

−5=f(96, 65)− f(96, 70)

−5=

121− 125

−5= 0.8

Averaging the two values results in fH(96, 70) ≈ 0.9%. This means that when therelative humidity is 70% and the actual temperature is 96◦F , the heat indexincreases by about 0.9◦F for every percentage increase in relative humidity.

Functions of three or more variables

Partial derivatives of functions of more than two variables can be de�ned and computedin a way similar to the case of functions of two variables. For example, fx(x, y, z) for afunction of three variables is de�ned by

De�nition 2.

fx(x, y, z) = limh→0

f(x+ h, y, z)− f(x, y, z)h

Thus, to compute fx(x, y, z), we regard the variables y and z as constants, and forfy(x, y, z), we consider x and z as �xed, and so on.

Example 3. (1) Given that f(x, y, z) = xy − x2y3z4, �nd fx, fy, dz.

(2) Find∂z

∂xand

∂z

∂yfrom the following expression

x2z3 + xzy3 = x3 + y3

Solution 3. (1) f(x, y, z) = xy − x2y3z4

fx = y − 2xy3z4, fy = x− 3x2y2z4, fz = −4x2y3z3

(2) x2z3 + xzy3 = x3 + y3. We apply implicit di�erentiation, and �rst �nd∂z

∂x.

2xz3 + 3x2z2∂z

∂x+ zy3 + xy3

∂z

∂x= 3x2

(3x2z2 + xy3)∂z

∂x= 3x2 − 2xz3 − zy3

∴∂z

∂x=

3x2 − 2xz3 − zy3

(3x2z2 + xy3).

We now �nd∂z

∂y.

3x2z2∂z

∂y+ 3xy2z + xy3

∂z

∂y= 3y2

(3x2z2 + xy3)∂z

∂y= 3y2 − 3xy2z

∂z

∂y=

3y2 − 3xy2z

(3x2z2 + xy3)



1.1.2 Higher Order Partial Derivatives

We can �nd higher order partial derivatives of functions of several variables similar tothe case of single-variable functions. For instance, in the case of a function of twovariables z = f(x, y), we have

(fx)x = fxx =∂

∂x

(∂f

∂x

)=∂2f

∂x2

(fx)y = fxy =∂

∂y

(∂f

∂x

)=

∂2f

∂y∂x

(fy)x = fyx =∂

∂x

(∂f

∂y

)=

∂2f

∂x∂y

(fy)y = fyy =∂

∂y

(∂f

∂y

)=∂2f

∂y2

The partial derivatives fxy and fyx are often called mixed partial derivatives becausewe are di�erentiating with respect to di�erent variables. In the expressions above, theorder of di�erentiation is important. The expressions in the brackets need to be

di�erentiated �rst. For example, fxy =∂2f

∂y∂xmeans that we �rst di�erentiate f with

respect to x and then di�erentiate the resulting expression with respect to y. Also note

that the order of the variables x and y in the notations fxy and∂2f

∂y∂xare reversed.

Example 4. Find the second partial derivatives of

f(x, y) = x2y + sin(3x)− 5xy3

Solution 4.

fx = 2xy + 3 cos(3x)− 5y3, fy = x2 − 15xy2

fxx =∂

∂x(2xy + 3 cos(3x)− 5y3) = 2y − 9 sin(3x)

fxy =∂

∂y(2xy + 3 cos(3x)− 5y3) = 2x− 15y2

fyx =∂

∂x(x2 − 15xy2) = 2x− 15y2

fyy =∂

∂y(x2 − 15xy2) = −30xy

In the example above we notice that fxy = fyx. This is not always the case but it is thecase for most functions that we encounter in real life. The following theorem providesconditions under which we can expect the mixed partial derivatives to be equal.

Theorem 1 (Clairaut's Theorem). If the function f is de�ned on a disk D containingthe point (a, b), and if fxy and fyx are both continuous on D, then

fxy(a, b) = fyx(a, b).



Remark 2. Partial derivatives of order higher than two can also be de�ned. For example

fxyx = (fxy)x =∂

∂x

(∂2f

∂y∂x

)=

∂3f

∂x∂y∂x

fyxx = (fyx)x =∂

∂x

(∂2f

∂x∂y

)=

∂3f

∂x2∂y

All the partial derivatives considered so far can be extended to a function of any numberof variables. Clairaut's Theorem can also be extended to any order of mixed partialderivatives provided the number of times we �nd the partial derivative with respect to avariable is the same in each mixed partial derivative. For instance, fxxyyyz = fxyyxzy,provided the mixed derivatives are continuous functions.

Example 5. Find fyyxz if f(x, y, z) = cos(z + xy) + xyz

Solution 5.

fy = −x sin(z + xy) + xz

fyy = −x2 cos(z + xy)

fyyx = −2x cos(z + xy) + x2y sin(z + xy)

fyyxz = 2x sin(z + xy) + x2y cos(z + xy)


math 355: calculus of several variables · 1|partial di erentiation 1.1partial derivatives de...

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