Engineering Mathematics IIIBET 2543

Chapter 2Partial Derivative

Partial Derivative

Partial Derivatives Increments and Differentials Chain Rules Local Extrema Absolute Extrema

Partial Derivative

dxd of instead used is operator ation,differenti partialIn

sderivative partial ofconcept the toleads This fixedremain ablesother vari

the while variablesits of one .r.t.function w theof change of rate in the interested are we variable,one than more offunction aFor

derivative itsby drepresente is function theof change of rate the variable,one offunction aFor

x

Partial Derivatives

Example

31)1(2)0(3)1,0( Hence

123010230),(

obtain to torespect with ),( atedifferenti weconstant, a as x regardingBy

11)1)(0(6)0(4)1,0( Hence164

001064),( obtain to to

respect with ),( atedifferenti weconstant, a as regardingBy

132),( (i)

if )10( and , )10( Find

2

22

222

y

y

x

x

yx

f

yxyxyxf

yyxf

fxyxxyxyxf

xyxfy

yxyyxxyxf

,f,f

Partial Derivatives

Example

0432xz2

obtain weHence .z3xz and z2

xz that have We

04)2(

0)4()()(

obtain to respect toith equation w theof sidesboth partially atedifferenti We

054

by implicitly defined is isy and x variables twooffunction a is z if y,z and z Find

2222

232

322

22

3222

3222

xzy

xzzyxzzx

xz

xz

xzy

xzyxz

xzx

yzxx

zxyx

zxx

x

yzzxyzx

x

Partial Derivatives

22

22

342)2(

xz

obtain weside, handright therest to theand side handleft the toz containing termCollecting

zyzxyxz

x

Partial Derivatives

2

2

2

2

2

2

yx

yx

)(y

)(x

)(y

)(x

by denoted areThey y).f(x, of sderivative partial second theare

),( and ),( of sderivative Partial . variables twoof functions also

are ),( and ),( then , variables twooffunction a is )( If

yf

yf

yff

xxf

yf

xff

xyf

xf

yff

xf

xf

xff

yxfyxf

yxfyxfx,yf

yyy

yxy

xyx

xxx

Partial DerivativesExample

yxxyxy

fy

f

xxyxyxx

fx

f

xxyyxxyy

fy

f

yxyyxxyx

fx

f

f

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f

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yx

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2422

32422

3233

2333

42243233432

432,

6)3()(

46)3()(

46)42()(

122)42()(

are of derivative second The

3)( , 42)(

are of derivativefirst The

),( if ,, Find

Partial Derivatives

xy

xyxy

xyy

xy

xyxy

xyx

xy

yyyxxyxx

exxy

xeyxe

eyxy

yxf

eyxy

yeyxe

eyxx

yxf

eyxy

ffff

)1(

)()10(

])[(),(

)1(

)()01(

])[(),(

)(given if

Find

2

2

,,,

Example

Partial Derivatives

xy

xyxy

xyxxx

eyxyy

yeyxyey

eyxyx

fx

yxf

)2(

)1()00(

])1[()(),(

32

2

2

xy

xyxy

xyxxy

exyyxyx

xeyxyeyx

eyxyy

fy

yxf

)22(

)1()20(

])1[()(),(

22

2

2

Partial Derivatives

xy

xyxy

xyyyy

exyxx

xexxyex

exxyy

fy

yxf

)2(

)1()00(

])1[()(),(

32

2

2

xy

xyxy

xyyyx

exyyxyx

yexxyexy

exxyx

fx

yxf

)22(

)1()20(

])1[()(),(

22

2

2

♣

Increment and Differential

yyxfyxfyyxfxyxfyxfyxxf

yyxfyyxfyxf

xyxfyxxfyxf

yyxfyyxxfyxf

xyxfyyxxfyxf

yxfyyxxfzzzy

yxxx,yPyxfz

y

x

yx

yy

xx

),(),(),( ),(),(),(

implieswhich

),(),( ),( ,),(),(),( obtain we

),(),(limit),(

,),(),(limit),(

s,derivative partial of definition theFrom),(),(

then,in increasean is and in increasean is ,in increasean is If surface. on the )(point a

consider uslet and ),(fucntion agiven are that weSuppose

0

0

Increment and Differential

0.06 (-0.02)-2)0.01(6Δz

obtain we-0.02,y and 0.01x and 2,y 1, xngSubstituti

obtain weHence

and 6 have We

)(1.01,1.98 to(1,2) from changes )( and 3),(

if derivative partial usingby zfor ion approximatan Find 2

xfyxf

x,yxyxyxfz

yx

Example

Increment and Differential

12.0 12(14)0.0103)14(10)(-0.12(10)0.02

),,(),,(),,(V

in volume. change theeapproximat top.d Use10.01. and 11.97, 14.02, to10 and 12 14,

from changebox r rectangula a of cm)(in dimension theSuppose

zxyyxzxyz

zzyxVyzyxVxzyxV zyx

Example

Chain Rule

,

,

thenexist,h and g f, of sderivative partial with )( and )( ),( If : Theorem

variable.one of functionsfor ruleschain theofextension an is variablesone than more of functionsfor ruleschain The

yv

vz

yu

uz

yz

xv

vz

xu

uz

xz

x,yhvx,yguu,vfz

Chain Rule

variables twooffunction for diagram threeThe

below. figurein shown as is above mfor theore diagram threeTherule.chain heremember t us help toused becan diagram threeThe

Chain Rule

yvxyu

yxvyuxv

vz

xu

uz

xz

xyvyx

xvxy

yuy

xu

y

vvzu

uz

vuz

yzxzyxvxyuv

sin43

)sin2(2)(3

obtain weHence

cosy , sin2 ,2 ,

,sinx vand xyu fromSimilarly

2 and 3

obtain we, From

. and rulechain a Use.sin,and,uz Suppose

22

22

22

22

2

23

2233

Example

Chain Rule

2sin6

cos)sin(2)(6

cos26

sin43

sin)sin(4)(3

sin43

obtain we,sin and ngSubstituti

cos26

)cos(2)2(3

Conts

453

2222

22

2362

2222

22

22

22

22

yxyx

yxyxxyxy

yvxxyuyz

yxyx

yxyxyxy

yvxyuxz

yxvxyu

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yxvxyuyv

vz

yu

uz

yz

Chain Rule

. variables threeof functionsfor diagram threeThe

. and ,, offunction a iseach r and , , whereas, and , , offunction a is that suppose example,an As

rules. theseformulate help todconstructe becan diagrams treeand les, variabofnumber any of functions composite toapplied becan ruleChain

zyxvurvu

Chain Rule

16)320(3

)12(9)(6)43(12

9612

)3(3)2(3)6(2

dtdw

obtain wediagram, tree theFrom

and ,12 ,43 with 3

if find torulechain the Use

2

232

2

2

322

t-tt

ttttt

ytztx

tyztx

dtdz

zw

dtdy

yw

dtdx

xw

tztytxyzxw

dtdwExample

Chain Rule

35512

35)512(

),(),(

obtain we,1543 From

1543by defined

implicitly is offunction a as if , Find

),(),(

then , variableone offunction a as function abledifferenti a defines implicitly 0) If : Theorem

4

2

4

2

35

35

yx

yx

yxFyxF

dxdy

x-xy- y

x-xy-y

xydxdy

yxFyxF

dxdy

xy F(x,y

y

x

y

x

Example

Local Extrema

function. a of values minimumor maximumeither refer to extremum wordThe

point. saddleor point minimum point, maximum- :iespossibilit three

has )(graph of point) stationary(or point Critical

x,y fz

Local Extrema

R.in )(every for )( )(such that D,Rregion a exists thereif f,function theof minimum local a is )((ii)

R.in )(every for )( )(such that D,Rregion a exists thereif f,function theof maximum local a is )( (i)

Din b)(a, that and D,domain ain defines variables twooffunction a is )( Suppose

:

x,yx,yfa,bfa,bf

x,yx,yfa,bfa,bf

x,yfz

Definition

Local Extrema

Point (a,b,f(a,b)) is a (local) maximum

Local Extrema(i) Minimum point

Point (a,b,f(a,b)) is a (local) minimum

Local Extrema

Point (a,b,f(a,b)) is a saddle point

Local Extrema

0b)G(a, if made becan sconclusion No (iv)0b)G(a, ifpoint saddle a is )),(,,( (iii)

0),( and 0b)G(a, if minimum local a is ),( (ii)0),( and 0b)G(a, if maximum local a is ),( (i)

Then

)],([),(),(),( ),(

),( ),(b)G(a,

Let R.in point critical a is b)(a, and Rregion aon sderivative partial second continuous

has that variables twooffunction a is )( that Suppose Test) Derivative (Second Theorem

2

bafbabafbafbafbaf

bafbafbafbafbaf

bafbaf

x,yf

xx

xx

xyyyxxyyxy

xyxx

Local Extrema

xyyxyxf

yxyxyxf

yxyyxf

111),( )iii(

16),( )ii(3

),( (i)

:functions following theof points stationary heidentify t and Locate

44

23

Example

Local Extrema

40)2(2)G(

2 0 2 )G( and derivative second theFind : 3 Step

1y 01

0 02

uslysimultaneo 0f and 0f Solve :2 Step

1 , 2

sderivative partialfirst theFind : 1 Step3

),( )i(

2

2

yx

2y

23

yyfffx,y

yfffx,y

y

xx-

y fxf

yxyyxf

xyyyxx

yyxyxx

x

point maximum a is )32(0,-1, , Therefore

3210

3)1()1,0(

02)1,0( 04)1(4G(0,-1) (0,-1)At

point saddle a is )32(0,1,- Therefore,

3210

31)1,0(

04)1(4G(0,1) (0,1)At (0,-1) and (0,1)point Test the : 4 Step

3

f

f-

f

-

xx

Absolute Extrema

region. in the minimum absolute theis minimum local then theregion, in thepoint saddle no and

maximum local no is thereand minimum local aonly assumes y)f(x, If (ii)region. in the maximum

absolute theis maximum local then theregion, in thepoint saddle no and minimum local no is thereand maximum local aonly assumes )( If (i)

R.region ain continuous is )( Suppose :

region.given any on valueestremum theisfunction afor estremum Absolute :

region.given only on y)f(x, of valueextremum thefind touseful more isIt ns.applicatio

practicalmost for enough not is extremum local thefindingClearly

x,yfx,yfTheorem

Definition

Absolute Extrema

(2,2)point critical theonly test weR,region theoutside is (0,0) Since (2,2). and

(0,0) points criticalobtain weusly,simultaneo equations theSolving063 and 063

have weThus

usly.simultaneo 0),( and 0),(such that )( Findpoints. critical Find : 1 Step

.0,0:),(region in the 6),( of extremum absolute theFind

22

33

xyyx

yxfyxfx,y

yxyxRxyyxyxf

yx

Example

Absolute Extrema

R.in minimum absolute theis -8(2,2)

minimum local the thereforeR,in maximum local no is thereSince

minimum. local a is 8)2)(2(622)2,2(

Therefore 012)2,2( and 0108)6()12(21G(2,2)

point. critical heclassify t and G(2,2) Compute : 3 Step

12 )2,2( thus6 )(

6 )2,2( thus6 )( 12)22( thus6)(

(2,2) and (2,2) (2,2), Compute : 2 Step

33

2

f

f

f--

fyx,yf

fx,yf, fx x,yf

fff

xx

xyyy

xyxy

xxxx

yyxyxx

Any questions?