MAT495_CHAPTER_2

Part 1 PARTIAL DERIVATIVES MAT 295

23

Chapter 2

Partial

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

Define partial derivatives

Interpret partial derivatives

State and apply the rules of partial derivatives

2.1 Introduction

Chapter 2 will continue the discussion on functions of two variables. In the

remainder of this chapter the differentiation for functions of more than one

variable will be looked into. The process of differentiation requires one of the

variables to be changed at a time, while the remaining variable(s) are held

fixed. In other words, what should be done if only one or more than one of the

variables need to be changed?

2.2 Partial derivatives

Functions of several variables are functions of 2 or more variables. The

symbol ‘ ’ is used for differentiating function of more than one variable. In

general, if ),( yxfz the notation used to represent the partial derivatives are:

fDfDfx

zyxf

xx

ffyxf xxx

11),(),(

fDfDfy

zyxf

yy

ffyxf yyy

22),(),(


24

Definition

If ),( yxfz , then the first partial derivatives of f with respect to x and y are

the functions xf and yf respectively defined by

x

yxfyxxfyxf

xx

),(),(lim),(

0

y

yxfyyxfyxf

yy

),(),(lim),(

0

provided the limits exist.

This definition indicates that if ),( yxfz , then to find xf , just consider y

constant and differentiate with respect to x. Similarly, to find yf , consider x

constant and differentiate with respect to y.

Example 1

Use the definition of partial derivatives to find x

f

and

y

f

given

yxyxf 2),(

x

yxyxx

x

yxfyxxf

x

f

x

x

)()(lim

),(),(lim

22

0

0

x

xxx

x

yxyxxxx

x

x

2lim

)(2lim

0

222

0

Steps : Differentiate functions of two variables

To find xf , assume y as a constant and

differentiate ),( yxf w.r.t. x

To find yf , assume x as a constant and

differentiate ),( yxf w.r.t. y

Solution


25

x

xxx

f

x

2

2lim0

1

lim

lim

lim

),(),(lim

0

22

0

22

0

0

y

y

y

yxyyx

y

yxyyx

y

yxfyyxf

y

f

y

y

y

y

Example 2

Use the definition of partial derivatives to find ),( yxf x and ),( yxfy if

23),( yxyxf .

x

yxxyxxyxyxyx

x

yxyxxxxxx

x

yxyxx

x

yxfyxxf

x

f

x

x

x

x

2332222223

0

2323223

0

2323

0

0

)()(33lim

)()(33lim

)(lim

),(),(lim

22

22222

0

22222

0

322222

0

3

)(33lim

)(33lim

)()(33lim

yx

xyxxyyx

x

xxyxxyyx

x

xyxxyxyx

x

x

x

y

yxfyyxf

y

f

y

),(),(lim

0

y

yxyyx

y

2323

0lim

Solution


26

yx

yyx

y

yyyx

y

yxyyx

y

yxyxyyxyx

y

yxyyyyx

y

f

y

y

y

y

y

3

3

0

3

0

233

0

2323323

0

23223

0

2

2lim

2lim

2lim

2lim

2lim

Use the definition of partial derivatives to find x

f

and

y

f

given xyxyxf ),(

2.3 Interpretation

The partial derivatives of a function of two variables, ),( yxfz , have a

useful geometric interpretation. If 0yy , then ),( 00 yxfz represents the

curve formed by intersecting the surface ),( yxfz with the plane 0yy .

Therefore,

x

yxfyxxfyxf

xx

),(),(lim),( 0000

000

represents the slope of this curve at the point ),(,, 0000 yxfyx . Note that

both the curve and the tangent line lie in the plane 0yy . Similarly,

y

yxfyyxfyxf

yy

),(),(lim),( 0000

000

represents the slope of this curve given by the intersection of ),( yxfz and

the plane 0xx at the point ),(,, 0000 yxfyx .

Warm up exercise


27

Precisely, the values of x

f

and

y

f

at the point 000 ,, zyx denotes the

slopes of the surface in the x- and y-directions, respectively.

fx is the slope parallel to x-axis fy is the slope parallel to y-axis

Figure 2.1

Figure 2.2

Note:

The curve above line L1 has slope fx(a, b) at point P

The curve above line L2 has slope fy(a, b) at point P

Example 3

Find the slopes of the surface given by

22 )2()1(1),( yxyxf

at the point (1, 2, 1) in the x-direction and in the y-direction.

L1

z

y

x

L2 (a, b, 0)

P(a, b, f(a, b))

z = f(x, y)

z

L1

y x

(a, b,f(a,b))

fx = f(a,b)

L2

y x

(a, b,f(a,b))

fy= f(a,b)

z


28

The partial derivatives of f with respect to x and y are

)1(2),( xyxfx and )2(2),( yyxfy .

At the point (1, 2, 1), the slopes in the x- and y-directions are

0)11(2)2,1( xf and 0)22(2)2,1( yf .

Example 4

If 22 24),( yxyxf , find )1,1(xf and )1,1(yf and interpret these

numbers as slopes.

Compute the partial derivatives w.r.t x and y.

xyxfx 2),( and yyxfy 4),(

Evaluate the partial derivatives at the given points.

2)1,1( xf and 4)1,1( yf

2)1,1( xf is the slope of the tangent line to the intersection between

parabola 22 xz and the vertical plane y = 1 at the point (1, 1, 1).

4)1,1( yf is the slope of the tangent line to the intersection between

parabola 223 yz and the plane x = 1 at the point (1, 1, 1).

Partial derivatives can also be interpreted as rates of change, no matter

how many variables are involved. If z = f(x,y), then x

z

represents the rate

of change of z with respect to x when y is fixed. Similarly, y

z

represents

the rate of change of z with respect to y when x is fixed.

Example 5

The area of a parallelogram with adjacent sides a and b and included

angle is given by sinabA as shown in Figure 2.3.

Solution

Solution


29

a) Find the rate of change of A with respect to a for a = 10, b = 20 and

6

.

b) Find the rate of change of A with respect to for a = 10, b = 20 and

6

.

Figure 2.3

a) To find the rate of change of the area with respect to a, hold b and

constant and differentiate with respect to a to obtain

sinb

a

A

106

sin20

a

A

b) To find the rate of change of the area with respect to , hold a and b

constant and differentiate with respect to to obtain

cosab

A

31006

cos200

A

If xyxyxf ),( , find )1,1(xf and )1,1(yf and interpret these numbers

as slopes.

Solution

sinabA sina

b

a

Warm up exercise


30

2.4 Partial Derivatives of Basic Functions

In general if f is a function of two variables and suppose we want to find the

effect of changing the independent variable x to f, then y will be kept fixed.

Then we are really considering f as a function of a single variable x. In this

case the process of finding the partial derivatives of f w.r.t. x is similar to

finding an ordinary derivative.

Table 2.1 consists of ordinary derivatives of some basic functions. However,

the symbol “ ” is used in place of “d” to indicate partial differentiation.

Differentiating a function

1.

constant a is c w here

,0dx

dc

9. xsecxtan 2

dx

d

2.

constant a is n w here

,dx

d 1 nn nxx

10. xtan.xsecxsec dx

d

3. x

xln1

dx

d 11. ))(cot(coscos

dx

dxxecxec

4. elog.x

xlog bb1

dx

d 12. xeccosxcot 2

dx

d

5. xx ee dx

d 13.

2

1

1

1

dx

d

x

xsin

6.


, lndx

dccc xx

14.

2

1

1

1

dx

d

x

xcos

7. xcosxsin dx

d 15.

2

1

1

1

dx

d

xxtan

8. xsinxcos

dx

d

Table 2.1

Steps : Differentiate functions of two variables

Identify function : ),( yxf

Identify changed variable : x or y

Consider the function as a function of a single variable

Differentiate the function with respect to the identified variable


31

Example 6

Find fx and fy for the given functions:

a) yyxf ),( c) yxyxf 4),(

b) 23),( xyxf d) xyyxf 3),(

a) 0

y

xfx 1

y

yfy

b) xxx

fx 63 2

03 2

x

yfy

c) yxyxx

fx34 4

44 xyx

yfy

d) xyx

fx 3

xy

yfy 3

y

xy

yxy

xyx

3

32

1

332

1

3

2

1

2

1

2

1

x

xy

xxy

xyy

3

32

1

332

1

3

2

1

2

1

2

1

xy

y

xy

yfx

32

3

32

3

xy

x

xy

xfy

32

3

32

3

Example 7

Find x

f

and

y

f

for the given functions:

a) xyxf sin),( c) )ln(),( 4yxyxf

b) yeyxf 3),( d) )8tan(),( xyyxf

a) xxx

fx cossin

0sin

x

yfy

Solution

Solution


32

b) 03

y

x ex

f yyy ee

yf 33 3

c) yxyxx

fx34 4ln

44 xyx

yfy

d) xyx

fx 3

xy

yfy 3

y

xy

yxy

xyx

3

32

1

332

1

3

2

1

2

1

2

1

x

xy

xxy

xyy

3

32

1

332

1

3

2

1

2

1

2

1

xy

y

xy

y

32

3

32

3

xy

x

xy

x

32

3

32

3

Find xf and yf for the given functions.

(i) xyxf 3),( (iii) xyyxf cos),(

(ii) 223),( yxyxf (iv) )ln(),( 2xyyxf

2.5 Rules of Partial Derivatives

Before we proceed further, let us recall the rules of ordinary differentiation as

illustrated in Table 2.2. These rules for finding ordinary derivatives may also

be used to find partial derivatives.

Warm up exercise


33

1. 0cdx

d ; c is a constant 5.

dx

dgxfxg

dx

dfxgxf

dx

d )()()()(

2. 1 nn nxxdx

d 6.

2)(

)()(

)(

)(

xg

dx

dfxfxg

dx

df

xg

xf

dx

d

3. dx

dfxf

dx

d)( 7.

dx

dg

dx

dfxgf

dx

d)(

4. dx

df

dx

dfxgxf

dx

d )()(

Table 2.2

Table 2.3 illustrates the derivatives of a function with chain rule.

Differentiating a function with Chain rule (when u is a function of x)

1.

constant a is n w here

,)(dx

d.)(

dx

d 1 uunu nn 8. )u().u(sec)utan(

dx

d

dx

d 2

2. )u(.)u(

)uln(dx

d1

dx

d 9. )(

dx

d.)tan().sec()(

dx

duuuusex

3.

)(dx

d.log.

)(

1

)(logdx

d

ueu

u

b

b

10.

)()).))(cot(((cos

)(cosdx

d

udx

duuec

uec

4. )u(dx

d.ee )u()u(

dx

d 11. )(

dx

d).(cos)cot(

dx

d 2 uuecu

5.


, (u)dx

d. ln

dx

d )()( ccc uu

12. )u(.

)u(

)u(sindx

d

1

1

dx

d

2

1

6. )u().ucos()usin(dx

d

dx

d 13. )(

dx

d.

)(1

1)(cos

dx

d

2

1 u

u

u

7. )u().usin()ucos(dx

d

dx

d 14. )(

dx

d.

)(1

1)(tan

dx

d2

1 uu

u

Table 2.3

Again for functions of two or more variables, the symbol “ ” is used in place

of “d”.

Besides using these differentiation rules for algebraic operations, the chain

rule may also be used to differentiate composite functions.


34

Example 8

Let yxyxf 7),( and xyeyxg 2),( then compute

a) x

f

,

x

g

,

y

f

and

y

g

c) )(fg

x

and )(fg

y

b) )( gfx

and )( gf

y

d)

g

f

x and

f

g

y

a) Taking y as constant

7

x

f and xyye

x

g 22

Taking x as constant

1

y

f and xyxe

y

g2

b) Apply sum and difference rule

xyye

x

g

x

fgf

x227

)(

xyxe

y

g

y

fgf

y

221

)(

c) Apply product rule

xyxy yeyxe

x

gyxfyxg

x

ffg

x22 2)7()7(

),(),()(

)2()7()1(

),(),()(

22 xyxy xeyxe

y

gyxfyxg

y

ffg

y

Steps : Using rules of partial derivatives

Determine the algebraic operation

Select the appropriate rule

Differentiate the functions by holding all variables constant except for the changed variable

Solution


35

d) Apply quotient rule

),(

),(),(

2 yxg

x

gyxfyxg

x

f

g

f

x

22

22

)(

2)7(7xy

xyxy

e

yeyxe

22

2

)(

2)7(7xy

xy

e

yyxe

xye

yxy 22147

),(

),(),(

2 yxf

y

fyxgyxf

y

g

f

g

y

2

22

)7(

)1()7(2

yx

eyxxe xyxy

2

22

2

2

)7(

1214

)7(

172

yx

xyxe

yx

yxxe

xy

xy

Example 9

Given 42),( yxyxf , find

a) ),( yxfx c) ),( yxfy

b) )1,2(xf d) )1,2(yf

a) 42),( xyyxfx

b) 324),( yxyxfy

c) 4)1)(2(2)1,2( 4 xf

d) 16)1)(2(4)1,2( 32 yf

Solution


36

Example 10

Find x

z

and

y

z

a) 2345 102 yxyxz

b) yxyxz seclnsin 5

c) xyez x 2cos2

d) xy

yxez yx 12)tan( 5

a) 2345 102 yxyxz

2345 102 yxyxxx

z

224

224

2345

310

30010

102

yxx

yxx

yxxx

yx

xx

2345 102 yxyxyy

z

yxy

yxy

yxyy

yy

xy

33

33

2345

24

2040

102

b) yxyxz seclnsin 5

yxx

yxx

yxyxxx

z

seclnsin

seclnsin

5

5

yxyx

xyxxyx

yx

xyxx

yx

xyxx

sec5lncos

)0(sec5)0(sinlncos

secseclnsinlnsin

4

54

55

yxy

yxy

yxyxyy

z

seclnsin

seclnsin

5

5

Solution


37

yyxy

x

yyxyy

xy

yy

xyxy

yy

xyxyy

z

tansecsin

)tan(secsec)0(1

sinln)0(

secseclnsinlnsin

5

5

55

c) xyez x 2cos2

)2sin(22cos2

)2()2sin(2cos2

2cos2cos

2cos

22

22

22

2

xyyexye

yxyexye

xyx

exyex

xyexx

z

xx

xx

xx

x

xyy

exyey

xyeyy

z

xx

x

2cos2cos

2cos

22

2

)2sin(2

)2()2sin(2cos)0(

2

2

xyxe

xxyexyy

z

x

x

d) xy

yxez yx 12)tan( 5

12

152

5

5

2))((sec

12)tan(

12)tan(

xyx

yxx

ee

xyxyx

xe

x

xyyxe

xx

z

yxyx

yx

yx

yxyx

ee

xy

y

yx

ee

yxyyxee

yxyx

yxyx

yxyx

25

2

25

2

22

152

1

2

1)(sec

)(2

1)(sec

)((222

1)(sec

xyyxe

yy

z yx 12)tan( 5


38

12

152

5

2))((sec

12)tan(

xyy

yxy

ee

xyyyx

ye

yy

z

yxyx

yx

25

42

25

42

242

152

1

22

5)(sec

)(22

5)(sec

)()(522

1)(sec

xyyx

yee

xy

x

yx

yee

xxyyyxee

yxyx

yxyx

yxyx

Example 11

Given 322 )sin(),,( zxyzyxzzyxfw find dy

dw

x

w,

and

z

w

.

Taking y and z as constant

32

32

2 zxy

zyz

x

w

Taking x and z as constant

32

32 )cos(2

zxy

xyzzyyz

y

w

Taking x and y as constant

32

2222

2

3)cos(

zxy

zxyzyyx

y

w

Example 12

Show that 4224 CyyBx2Axu satisfies the function u4y

uy

x

ux

Given 4224 CyyBx2Axu

u4y

yy

x

ux

Solution

Solution


39

23 Bxy4Ax4x

u

32 Cy4yBx4y

u

)Cy4yBx4(y)Bxy4Ax4(xy

uy

x

ux 3223

422224 Cy4yBx4yBx4Ax4

)CyyBxyBxAx(4 422224

u4

Given that xyxyyxyxf lncos5),( 43

(i) Find xyxyyxx

lncos5 43

(ii) Then compute )0,1(xf by substitute x=1 and y=0

(iii) Next find xyxyyxy

lncos5 43

(iv) Compute )0,1(yf by substitute x=1 and y=0

Example 13

Find pf and qf

i) )(lntan),( 1 pqqpf ii) )cot(),( 34 pqpqpf

i) )(lntan),( 1 pqqpf

)(ln)(ln1

12

pqppqp

f

)(1

)(ln1

12

pqppqpq

Solution

Warm up exercise


40

)(1

)(ln1

12

qpqpqp

f

ppq

1

)(ln1

12

)(ln)(ln1

12

pqqpqq

f

)(1

)(ln1

12

pqqpqpq

)(1

)(ln1

12

ppqpq

qpq

1

)(ln1

12

b) )cot(),( 34 pqpqpf

p

uv

p

vu

p

f

))((csc)cot(4 332433 qpqppqp ;

)(csc)cot(4 324333 pqqppqp

q

vqpuqpv

q

u

q

f

),(),(

)3)((csc 2324 pqpqp

)(csc3 3225 pqqp

Example 14

For 2322),( sresrf rs , find )1,0(rf and ).1,0(sf

322

322

232

4

22

2

2

2

srres

rsres

srr

er

f

rs

rs

rsr

1

)1()1()1,0( 2

rf

Solution


41

322

232

232

62

322

2

2

2

srsrse

ssrrse

srs

es

f

rs

rs

rss

2xy 2 3 22xye 2(x y )(3y )

2xy 2 2 32xye 6y (x y )

6

)1()1(6)1,0( 2

sf

Let rsrsrf 3),( 2 and rssrg cos),( .

(i) Find the partial derivatives rf and sf .

(ii) Find the partial derivatives rg and sg .

(iii) Then compute

a) ).( gfr

by using

r

fsrg

r

gsrfgf

r

),(),().(

b) ).( gfs

by using

s

fsrg

s

gsrfgf

s

),(),().(

Exercise 2

Partial Derivatives

1. Let ),( yxfz . Define ),( yxfx and ),( yxfy . Write other notations which can be

used to represent ),( yxfx and ),( yxfy .

2. Derive x

Z

and

y

Z

using the definition of partial derivatives (limits):

a) 25),( yxyxZ b) 222),( yxyxZ

3. If 22),( yxyxf find )4,3(xf and )2,0(yf

4. Given )32sin(),( yxyxZ , find )4,6(

y

Z

Warm up exercise


42

5. In each of the following, find x

f

and

y

f

.

a) 52),( yyxyxf e) )ln(),( 32 xxyyxf

b) )tan(),( xyeyxf xy f) )sin(),( 77 xyexyxf

c) 22

22

),(yx

yxyxf

g)

yx

xyyxf

3

3 )(sec),(

d) yxyxf 3cos2sin),( h) x

yyxyxf

52),(

6. Find all the first order derivatives for the following functions:

a) 423),( yxyxZ d) qpqpZ ln),(

b) sin),,( txetxU e) 22 4),( yxyxf

c) )32ln( zyxW f) 222 zyxU

7. Let 233),( yxyxf . Find:

a) ),( yxfx d) ),( yxfy

b) ),1( yfx e) )1,(xfy

c) )2,1(xf f) )2,1(yf

8. Let yez x sin2 . Find:

a) x

z

b)

y

z

9. If ),( yxxfz , show that zy

zy

x

zx

10. Find xf and yf .

a) yx

yxyxf

),( c)

x

yxyxf cos),( 3

b) xyexyxf y sec),( 33 d) )ysin(e)y,x(f xy 24

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MAT495_CHAPTER_2