MAT495_CHAPTER_5

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Part 1 PARTIAL DERIVATIVES MAT 295

67

Chapter 5

Total

differentialsAt the end of this chapter, students should be able to:

Define total differential

Apply total differential in words problem

5.1 Introduction

In chapter 5, the concepts of increments and differentials of a single variable

functions will be generalized to functions of two or more variables. Given a

function of a single variable )(xfy and a point ))(,( cfc on the curve, the

quantity

)()( 00 xfxxfy

is the change in yalong the curve )(xfy produced by the change x inx

Figure 5.1 Increment y in 2

x0+xx0

y

f(x0+x)

f(x0)

x

x

y

(c,f(c))

f(x)

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The differential of ywas defined as

dxxfdy )('

and represents the change inyalong the tangent line at cproduced by a change

in x.

55..22 Total DDiiffffeerreennttiiaall

Similar terminology is used for a function of two variables, ).,( yxfz That is,

x and y are the increments ofxand y, and the increment of zis z .

Definition

If ),( yxfz , the increment of zdenoted as z is given by

),(),( yxfyyxxfz

Note that the increment z depends on x, y, x and y , and is equal to the

change in zasxchanges by x and ychanges by y . Precisely, the increment in

z refers to the actual change in z when x and/or y changes. To be exact, z

represents the change in the value of z when ),( yx changes from ),( yx to

),( yyxx .

Figure 5.2: Increment z in3

x

y

y

x

f(x,y)

(x,y)

f(x+x, y+y)

z

(x+x, y+y)

z

x

y

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Definition

If ),( yxfz , the total differentialof zdenoted as dzand given by

dyyxfdxyxfdz yx ),(),(

or equivalently,

dyy

zdx

x

zdz

.

This formula represents the estimated change in z.

dzz

In other words, the estimated change in z may be evaluated by using partial

derivatives such that the total change in z (called total differential) is the

summation of the change in zdue to the change inx(keeping yconstant) and the

change in zdue to the change in y (keeping xconstant). The notations dzand

df can be used interchangeably since ),( yxfz .

The above definition can be extended to a function of three or more variables. For

instance, given the function ),,( zyxgw the differential in wdenoted as dw is

given by,

dzgdygdxgdw zyx .

Example 1

Find the differential df given xyyxyxf sincos),( .

Identify the variables and partial derivatives

Variables : xand y

Partial derivatives: xf and yf

Find partial differentiation

xyyfx coscos

Solut ion

Steps : Total differential

Identify the variables and partial derivatives Find partial differentiation Apply total differential

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70

xyxfy sinsin

Apply total differential

dyfdxfdf yx

dyxyxdxxyy )sinsin()cos(cos

Example 2

Find the differential dfgiven yxxyyxf 2)ln(),( .


Variables :xand y



xx

xyxy

fx 21

2)(1

11

1)(1

y

xxy

fy


dyfdxfdf yx

dyy

ydxx

x

12

1

Example 3

Find the increment and total differential of yxyyxf 22),( .

Increment (actual change) in f:

),(),( yxfyyxxff

yxyyyxxxxyyyy

yxyyyxxyyf

222222

2222

)(2)(2

Solut ion

Solut ion

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yxyyxyxxyxxyxxyyxyyyyf

22

222222 )(2)(2)(2

22

2222

)(222

)(2)(2)(2

xxxxyyyxyxyx

yxyxxyxxyxxyyyy

Total differential (estimated change) in f:

dyy

fdx

x

fdf

xyfx 2

22 xyfy

dyxydxxydf 222

Let 22),( qpqpw .

(i) Find the increment (or actual change) in w if (p,q) changes to

qqpp , .(ii) Write the formula for estimating the change in w using partial derivatives.

(iii) If (p,q) changes from (3,4) to (3.04, 3.98).

a) Calculate w.

b) What is pand q

c) Calculate dw and compare your answer with the value w. Whats

your observation?

55..33 AApppplliiccaattiioonnss

In measurement, the total differential is used in estimating fof a function fbased

on xof the independent variable x and yof the independent variable ysuch

that

yyxfxyxff yx ),(),(

Warm up exercise

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Example 4

Estimate by a differential the change in f if ),( yx changes from (32, 16) to

(35, 18) given that yxyxyxf 52

),( .


Variables : xand y



yyxfx

5

3

52

y

xxfy

2

152

Identify values ofx, y, xand y (if any)

)16,32(),( yx and )18,35(),( yyxx

33235 x 21618 y


yfxffyx

Solut ion

Steps : Total differential

Identify the variables and partial derivatives Find partial differentiation Identify values ofx, y, xand y (if any) Apply total differential

(32,16)

(35, 18)

x=3

y=2

zdz

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yy

xxxyyxf

2

1

5

2 52

5

3

Then substitute (32, 16) into the formula of total differential

)2(16

)32(

2

1)32()3(16)32)(16(

5

2 52

5

3

df

)44(245

43

4.30

5

230

Example 5

Use differential to estimate 3 102127

Step 1: Represent the problem into mathematical expression in the form of a

function

3),( yxyxf

3 102127)1021,27( f

Step 2: Find partial differentiation

x

yyxfx

2),(

3

3 23

),(

y

xyxfy

Step 3: Identify value ofx, y, x, and y.

For x find the closest to 27:

x= 25; x= 2725 = 2

For 3 y the closest to 1021:

y= 1000; y= 1021- 1000 = 21

110

10)1000,25(

xf

60

1

)100(3

5)1000,25(

yf

Solut ion

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Step 4: Apply total differential

dyfdxfdf yx

35.2

)21(60

1)2)(1(

In this case, )1021,27(f changes to )1000,25(f or

fyxfyyxxf ),(),(

fff )1000,25()1021,27(

dff )1000,25( ; dff

35.52

35.2)10(50(

35.2100025 3

Example 6

Given that 4kwy , where k is a constant. Calculate the approximate

percentage in yif wis increased by 3% and is decreased by 2.5%.

Given

4kwy

wwofw 03.0%3

025.0%5.2 of

4kyw

34 kwy

ywydy w

)025.0(4)03.0( 34 kwwk

44 1.003.0 kwkw

)1.003.0(4 kw

ykw 07.007.0 4

Therefore y is decreased by 7%.

Solut ion

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Let

4

),( yxyxf

(i) Find xf and yf .

(ii) Find xand yif the change in fis from (121, 16) to (125, 17),

(iii) Find xf and yf ifx= 121 and y= 16.

(iv) Use the total differential to find the approximate value of 4 17125 .

Exercise 5

Total Differential and its Application

1. Find the differential df is yxyxyxf ln),( .

2. Use the total differential to find the approximate value of 4 17125 .

3. Use the differential to estimate the change in yxyxf cos),( 2 from (2, 2)

to (2.1, 1.9).

4. Use the total differential to approximate the change in ),( yxf as ),( yx varies from

P to Q. Compare your estimate with the actual change in ),( yxf .

a) xxyxyxf 42),( 2 ; P(1, 2), Q(1.01, 2.04)

b) 21

3

1

),( yxyxf ; P(8, 9), Q(7.78, 9.03)

c)xy

yxyxf ),( ; P(-1, -2), Q(-1.02, -2.04)

d) xyyxf 1ln),( ; P(0, 2), Q(-0.09, 1.98)

5. Use the differential to estimate the change in 2xyz from its value at (0.5, 1.0)

to its value at (0.503, 1,004). Compare the error in this estimate with the distance

between the points (0.5, 1.0) and (0.503, 1.004).

Warm up exercise

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