CHAPTER 5PARTIAL DERIVATIVES

INTRODUCTIONSMALL INCREMENTS & RATES OF CHANGE

IMPLICIT FUNCTIONSCHAIN RULE

JACOBIAN FUNCTIONHESSIAN FUNCTIONSTATIONARY POINT

INTRODUCTION

• Consider the following functions where

are independent variables.

• If we differentiate f with respect variable , then we assume thati. as a single variableii. as constants

nxxxf ,...,, 21

nxxx ,...,, 21

ix

ix

nii xxxxx ,...,,,..., 1121

NotationIf • First order partial derivatives:

• Second order partial derivatives:

,, yxff

y

f

x

f

and

x

f

yxy

fy

f

xyx

fy

f

yy

fx

f

xx

f

2

2

2

2

2

2

Example 1Write down all partial derivatives of the following function

xyyxyxyxf ln2cos2, 233

Example 1Write down all partial derivatives of the following function

SolutionFirst order PD

xyyxyxyxf ln2cos2, 233

xyyxyxy

fx

yyyx

x

f

ln22sin43

2cos23

23

232

Second order PD

xyxyx

xyyxyxy

y

f

yy

fx

yxy

x

yyyx

x

x

f

xx

f

ln22cos86

ln22sin43

6

2cos23

3

23

2

2

2

23

232

2

2

Second order PD (mixed partial)

x

yyyx

x

yyyx

y

x

f

yxy

fx

yyyx

xyyxyxx

y

f

xyx

f

22sin49

2cos23

22sin49

ln22sin43

22

232

2

22

23

2

In example 1, we observed that

This properties hold for all functions provided that certain smoothness properties are satisfies.

The mixed partial derivative must be equal whenever f is continuous.

xy

f

yx

f

22

Example 2Write down all partial derivatives of the following functions:

yxxyxxeyxfii

xxyexyxfixy

y

2

342

sin,235,

Solution 2

340340

801210

3206310

235,

42

42

422

24

2

2

4224

342

yy

yy

yy

y

xexy

fxe

yx

f

exy

fxe

x

f

xexy

fxyxe

x

f

xxyexyxfi

Solution 2

xxyxxyyxxeexyxy

f

xxyxxyyxxeexyyx

f

xyxexy

f

yxyyxyxyyeexyx

f

xxyxexy

f

xyxyxyexyex

f

yxxyxxeyxfii

xyxy

xyxy

yx

xyxy

xy

xyxy

xy

2cos2sin2

2cos2sin2

sin

2cos2sin2

cos

2cos

sin,

222

222

332

2

222

2

222

2

SMALL INCREMENTS & RATES OF CHANGE

Notation for small increment is Let then

i. A small increment in z, is given by

Where are small increments of the stated variablesii. Rate of change z wrt time, t is given by

,,...,, 21 nxxxfz

.

nxxx ,...,, 21

z

nn

xx

zx

x

zx

x

zz

...2

21

1

t

x

x

z

t

x

x

z

t

x

x

z

t

z n

n

...2

2

1

1

Example 3

The measurements of closed rectangular box are length, x = 5m, width y = 3m, and height, z = 3.5m, with a possible error of in each measurement.

What is the maximum possible error in the calculated value of the volume, V and the surface, S area of the box?

cm10

1

Solution 3

Volume of rectangular box:Possible error of the volume:

xyzV

x

y

z

V

zz

Vy

y

Vx

x

VV

Solution 3

Possible error of the volume:

xyz

Vxz

y

Vyz

x

VxyzV

43000

1.0351.05.351.05.33

zxyyxzxyz

zz

Vy

y

Vx

x

VV

Solution 3

Surface Area of rectangular box:Possible error of the volume:

yzxzxyA 222

x

y

z

A

zz

Ay

y

Ax

x

AA

Solution 3

Possible error of the surface area:

yxz

Azx

y

Azy

x

AyzxzxyA

222222

222

4601.032521.05.32521.05.3232

222222

zyxyzxxzy

zz

Ay

y

Ax

x

AA

Example 4

The radius r of a cylinder is increasing at the rate of 0.2cms-1 while the height, h is increasing at 0.5cms-1. Determine the rate of change for its volume when r=8cm and h =12cm.

Solution 4Volume of cylinder:

Rate of change:

hrV 2

3

1

t

r

r

V

t

h

h

V

t

V

h

r

125.0

82.0

?,

hdt

dh

rdt

drt

V

(1)

(2)

Solution 4From (1)Differentiate partially wrt t:

Substitute in (2):

2

3

1

3

2r

h

Vrh

r

V

4.70

2.01283

25.08

3

1

2.03

25.0

3

1

2

2

rhrt

V

IMPLICIT FUNCTIONS

DefinitionLet f be a function of two independent variables x and y, given by

constant. To determine the derivative of this implicit function:LetHence,

ccyxf ,,

yz

xz

dx

dy

dy

dy

y

z

dx

xd

x

z

x

z0

dx

dy

., cyxfz

Example 5Assume that y is a differentiable of x that satisfies the given function. Find using implicit differentiation.

2sin

323

2

342

yxxyxxeii

xxyexi

xy

y

dx

dy

Solution 5(i) Let

Then,

Therefore ,

323 342 xxyexz y

xexy

z

xyxex

z

y

y

34

632

42

24

xex

xyxe

dx

dyy

y

34

63242

24

THE CHAIN RULE

DefinitionLet z be a function of two independent variables x and y, while x and y are functions of two independent variables u and v.

The derivatives of z with respect to u and v as follows: Hence,

dv

dy

y

z

dv

xd

x

z

v

z

du

dy

y

z

du

xd

x

z

u

z

Example 6Let , where and

Find and

yexyz x22 vvux 32 2 uuvy 32

u

z

v

z

Solution 6

yexyz x22 x

x

exy

z

yeyx

z

2

2

2

22

vvux 32 2

32

4

2

uv

x

uvu

x

uuvy 322

3

6

12

uvv

y

vu

y

Solution 6Therefore

3222232223

2222

322233223

322

22

22

322613222623222

32212128122422

uuv

xx

uuv

xx

euuveuuuvuvexuyey

v

y

y

z

v

x

x

z

v

z

euveuuvuvvexuvyey

u

y

y

z

u

x

x

z

u

z

JACOBIAN FUNCTION

DefinitionLet be n number of functions of n variables

nfff ...,,, 21 nxxx ...,,, 21

nnn

n

n

xxxff

xxxffxxxff

,...,,

,...,,,...,,

21

2122

2111

JACOBIAN FUNCTION

Jacobian for this system of equations is given by:

OR

n

n

nn

n

n

x

f

x

f

x

f

x

f

x

f

x

fx

f

x

f

x

f

J

21

22

2

2

1

11

2

1

1

n

nnn

n

n

x

f

x

f

x

f

x

f

x

f

x

fx

f

x

f

x

f

J

11

2

2

2

1

2

1

2

1

1

1

Example 7Given and , determine the Jacobian for the system of equation.

yeu x 2sin2 yev x 2cos2

Solution 7Given and , determine the Jacobian for the system of equation.

yeu x 2sin2 yev x 2cos2

y

yyyy

yeye

yeyey

v

y

ux

v

x

u

J

xx

xx

4cos42cos2sin42cos42sin4

2sin22cos2

2cos22sin2

22

22

22

22

INVERSE FUNCTIONS FOR PARTIAL DERIVATIVES

DefinitionLet u and v be two functions of two independent variables x and y. . . Partial derivatives and are given by:

yxfvyxfu ,,, ,,,v

x

u

y

u

x

v

y

Jxu

v

y

J

yu

v

x

Jxv

u

y

J

yv

u

x

Example 8Given and , Find and

yeu x 2sin2 yev x 2cos2

,,,v

x

u

y

u

x

v

y

Solution 8Given and , Find and

yeu x 2sin2 yev x 2cos2

,,,v

x

u

y

u

x

v

y

y

ye

v

y

y

ye

v

x

y

ye

u

y

y

ye

u

x

xx

xx

4sin2

2sin

4sin2

2cos

4sin2

2cos

4sin2

2sin

22

22

Example 9Let and Find and

2223 43,3 vuxyxyxz

u

z

v

z

.52 vuy

HESSIAN FUNCTION

DefinitionLet f be a function of n number of variables . Hessian of f is given by the following determinant:

2

2

2

22

1

12

2

2

22

22

12

12

1

2

21

22

21

12

n

n

nn

n

n

n

n

x

f

xx

f

xx

f

xx

f

x

f

xx

fxx

f

xx

f

x

f

H

nxxx ...,,, 21

HESSIAN FUNCTION

Hessian of a function of 2 variables:Let f be a function of 2 independent variables x and y. Then the Hessian of f is given by:

22

2

2

2

2

2

22

2

2

2

yx

f

y

f

x

f

y

f

xy

fyx

f

x

f

H

HESSIAN FUNCTION

Stationary PointDefinition Given a function . The stationary point of occurs when and

Properties of Stationary Point

22

2

2

2

2

2

22

2

2

2

yx

f

y

f

x

f

y

f

xy

fyx

f

x

f

H

yxff , yxff ,0

x

f0

y

f

HESSIAN FUNCTION

Properties of Stationary Pointa) If H<0, then stationary point is a SADDLE

POINT

b) If H>0a) MAXIMUM POINT if

b) MINIMUM POINT if

c) If H=0, then TEST FAILS or NO CONCLUSION

0,02

2

2

2

y

f

x

f

0,02

2

2

2

y

f

x

f

Example 10Find and classify the stationary points of

223 23, yxyxxyxf

Solution 10Find stationary point(s):

1022

22

10263

2632

2

yxyx

yxy

f

yxx

yxxx

f

Substitute (2) in (1)

Stationary points:

3

4,0

3

4,0

0430430263

2

2

xx

yy

yyyyyyy

3

4,

3

4,0,0

Find the Hessian function:

4662

2

2

66

2

2

2

2

2

xH

yx

fy

f

xx

f

Determine the properties of SP:

Point Hessian: Conclusion

SP is a maximum point

SP is a saddle point

0,0

3

4,

3

4

4662 xH

02

06

08

2

2

2

2

y

fx

fH

08 H