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Chapter 5 Partial differentiation. 5.1 Definition of the partial derivative. the partial derivative of f(x,y) with respect to x and y are. for general n-variable. second partial derivatives of two-variable function f(x,y). Chapter 5 Partial differentiation. - PowerPoint PPT Presentation
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5.1 Definition of the partial derivative
yxy
xyx
fy
f
y
yxfyyxf
y
f
fx
f
x
yxfyxxf
x
f
)(),(),(
lim
)(),(),(
lim
0
0
the partial derivative of f(x,y) with respect to x and y are
Chapter 5 Partial differentiation
for general n-variable
i
ninii
xi
n
x
xxxxfxxxxxf
x
xxxxfi
),...,...,(),..,...,(lim
),....,,,( 2121
0
321
second partial derivatives of two-variable function f(x,y)
yxxy
yyxx
fxy
f
x
f
yf
yx
f
y
f
x
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)( )(
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)(.....)(.........)(
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x
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n
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Chapter 5 Partial differentiation
5.2 The total differential and total derivative
..............
),...,( function variable-nfor
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]),(),(
[]),(),(
[
),(),(),(),(
),(),(
and
22
11
21
nn
n
dxx
fdx
x
fdx
x
fdf
xxxf
dyy
fdx
x
fdf
dfyx
yy
yxfyyxfx
x
yyxfyyxxf
yxfyyxfyyxfyyxxf
yxfyyxxff
fffyyyxxx
Ex: Find the total derivative of with respect to , given
that
2/121
2/12
2/12
)1(
3sin32
)1(
1332
)1(
1 ;3 ,32
x
xxx
xxyx
dx
df
xdx
dyx
y
fyx
x
f
Chapter 5 Partial differentiation
xyxyxf 3),( 2 x
xy 1sin
5.3 Exact and inexact differentials
If a function can be obtained by directly integrating its total differential, the differential of function f is called exact differential, whereas those that do not are inexact differential.
aldifferentiinexact existdoesnot ),( function
3 (2)
aldifferentiexact ),()1( (1)
yxf
ydxxdydf
xxyyxfdxyxdydf
Inexact differential can be made exact by multiplying a suitable function called an integrating factor
exist g(x) ),( suitable no (2),(1)for
-(2)---)(),(
-(1)---)(3),(33
3
yh
xgxyyxfxdydyy
fx
y
f
yhxyyxfydxdxx
fy
x
f
dyy
fdx
x
fydxxdydf
Chapter 5 Partial differentiation
Ex: Show that the differential xdy+3ydx is inexact.
Properties of exact differentials:
x
yxB
y
yxA
x
B
yx
f
y
A
xy
f
yxBy
fyxA
x
fdfdyyxBdxyxA
),(),(
),( and ),(),(),(
22
for n variables
Chapter 5 Partial differentiation
ipsrelationsh )1(2
1, pairs allfor
if exact, is ),.......,,( 211
nnjix
g
x
g
dxxxxgdf
i
j
j
i
in
n
ii
Ex: Show that (y+z)dx+xdy+xdz is an exact differential
czyxzyxf
y
g
z
g
z
g
x
g
x
g
y
g
xgxgzyzyxg
)(),,( inspection by
0 ,1 ,1
, ,),,(
321321
321
5.4 Useful theorems of partial differentiation
dyy
zdx
x
zdzyxzz
dzz
ydx
x
ydyzxyy
dzz
xdy
y
xdxzyxx
xy
xz
yz
)()(),(
)()(),(
)()(),(
relation cyclic 1)()()( relation yreciprocit )()(
0constant a is if 0constant a is if
])()()[()()(
1
zyxzz
yxzzz
y
x
x
z
z
y
x
y
y
x
dxxdzz
dzz
x
z
y
y
xdx
x
y
y
xdx
Chapter 5 Partial differentiation
5.5 The chain rule
du
dx
x
f
du
dx
x
f
du
dx
x
f
du
dx
x
f
du
df
uxxxxxf
du
dy
y
f
du
dx
x
f
du
dfdy
y
fdx
x
fdf
uyyuxxyxff
n
n
in
i i
iin
........
)( and ),....,,( variables manyfor
)( ,)( and ),(for
2
2
1
11
21
mju
x
x
f
u
f
uuuxxxxxff
j
in
i ij
miin
,...,2,1
),...,,( and ),....,,(
1
2121
Chapter 5 Partial differentiation
5.6 Change of variables
Ex: Polar coordinates ρ and ψ, Cartesian coordinates x and y, x=ρcosφ,
y=ρsinφ, transform into one in ρ and φ 2
2
2
2
y
f
x
f
2
2
22
2
2
2
2
22
2
2
2
2
2222
21
2/122222
11),()( and )(
cossin ,
sincos
cos ,
sinsin
)/(1
/ ),tan
sin ,cos)(
,
ggg
y
f
x
fyxf
y
f
yyx
f
xx
f
yxxx
ρ
φ
y
φ
yx
y
xy
xy
x
φ (y/x
yyx
x
xyx
),(),( gyxf
Chapter 5 Partial differentiation
5.7 Taylor’s theorem for many-variables functions
00 ,0
00
00
22
222
2
2
00
)],()[(!
1),(
:is variables twofor series sTaylor' full *
),(at evaluated be to are sderivative the all
,
]2[!2
1),(),(
:variables twofor
yxn
n
yxfy
yx
xn
yxf
yx
yyyxxx
yy
fyx
yx
fx
x
fy
y
fx
x
fyxfyxf
])3(16)3)(2(48)2(27[2!
1
)3(7)2(93),(
2/ ,2/ ,/
/ ,/
226
666
22222322
2
yyxxe
yexeeyxf
exyyeyxfyexxeyfeyxf
xyeeyfeyxfxyxyxyxyxy
xyxyxy
Ex: The Taylor’s expansion of f(x,y)=yexp(xy) about x=2, y=3.
5.8 Stationary points of many-variables functions
yyxxxyyyxx
yyxxxyyyxx
yyxxxyyyxx
xx
xyyy
xx
xyxx
yyxyxx
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f
ffy
f
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fyyfxfxyxfyxf
y
f
x
f
2
2
2
222
00
00
2200
or sign opposite have and if :point saddle (3)
and negative are and both if :maximum (2)
and positive are and both if :minimum (1)
)]()([2
1),(),(
grearrangin , ,
]2[!2
1),(),(
0 and 0
Chapter 5 Partial differentiation
two-variable function about point ),( 00 yx
Ex: has a maximum at the point , a minimum at and a stationary point at the origin whose nature cannot be determined by the above procedures.
Chapter 5 Partial differentiation
)exp(),( 223 yxxyxf )0,2/3()0,2/3(
Sol:
minimum a :)0,2/3( maximum a: )0,2/3(
point stationary edundetermin an: )0,0(
0 ,)2/3exp(2/33 ,)2/3exp(2/36)0,2/3( pointsat (2)
0)00(point at (1)
)exp()32(2
)exp()24(
)exp()6144( are sderivative second the
)0,2/3( ,)0,2/3( ,)0,0(at are points stationary the
0yor 00)exp(2
2/3or 00)exp()23(
2222
2223
2235
223
2242
xyyyxx
xyyyxx
xy
yy
xx
fff
fff,
yxxyxf
yxyxf
yxxxxf
xyxyxy
f
xxyxxxx
f
for a n-variable function at all stationary points
Chapter 5 Partial differentiation
),.......,( 21 nxxxf
signs mixed have seigenvalue all :point saddle (3)
zero than less are seigenvalue all 02
1 :maximum (2)
zero thangreater are seigenvalue all 02
1 :minimum (1)
2
1
2
1
2
1
2
1
and
rseigenvecto orthogonal and seigenvalue real hasit symmetric, and real is
2/ elements withmatrix a define
2
1)()( :expansion sTaylor'
2
2
2
2
2
0
rr
r
rr
r
rr
rr m
rmmmrmmr m
Trmr
mmm
r
Trr
rr
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T
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af
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eaxeeeMe
en nM
xMxfxx
fM
xxxx
fxfxff
i
i
xx
f allfor 0
Chapter 5 Partial differentiation
Ex: Derivative the conditions for maxima, minima and saddle points for a function of two variables, using the above analysis.
yyxxxyxyyyxxyyxxyyxx
yyxxxyxyyyxxyyxxyyxx
yyxxxyxyyyxxyyxx
yyxxyyxx
yyxxxyxyyyxxyyxxyyxx
xyyyxxyyxx
xyyyxxyyyx
xyxx
yyyx
xyxx
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ffff
fff fffffff
fffff
fffff
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ff
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222
222
2222
222
22
2
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0 4)()(0 if
sign mixed have seigenvalue two :point saddle (3)
4)-( )(
00 and 0 :maxima (2)
4)( and 0 , 0
positive and real are seigenvalue two :minima (1)
]4)()[(2
1
0))((0
Find the maximum value of the differentiable function
subject to the constraint
Lagrange undetermined multipliers method
Chapter 5 Partial differentiation
5.9 Stationary values under constraints
cyxg ),(
),( yxf
),(
points stationary theat and of values the and find 0
0
that such choose can we dependent, are and
multiplier edundetermin Lagrange :
0)()()(
0 and 0 maximize
cyxg
yxy
g
y
fx
g
x
f
dydx
dyy
g
y
fdx
x
g
x
fgfd
dyy
gdx
x
gdgdy
y
fdx
x
fdff
Ex: The temperature of a point (x,y) on a circle is given by T(x,y)=1+xy. Find the temperature of the two hottest points on the circle.
Chapter 5 Partial differentiation
2/12/1 ,2/1
2/32/1 ,2/1
(1) intoput and 2/1(2) and (1) from
-(3)---020)()1(
-(2)---020)()1(
(1)-----1),( constraint
min
max
22
22
22
Tyxxy
Tyxxy
xy
yxyxy
xyy
xyyxx
xyx
yxcyxg
the stationary points of f(x,y,z) subject to the constraints g(x,y,z)=c1, h(x,y,z)=c2.
0)(
0)(
0)(
z
h
z
g
z
fhgf
z
y
h
y
g
y
fhgf
y
x
h
x
g
x
fhgf
x
Ex: Find the stationary points of subject to the
following constraints:
Chapter 5 Partial differentiation
333),,( zyxzyxf
0),,( and 1),,( (ii)
1),,( (i)222
222
zyxzyxhzyxzyxg
zyxzyxg
2/13/2 1 ,023 ,023
0 and 0 ,0 (a)
:zero is and , of somefor :conditionanother :note *
3/1at occur points stationary
2/31 intoput 3/2
023)(
023)(
023)( (i)
2222
222
2
2
2
zyzyzzyy
zyx
zyx
zyx
zyxzyx
zzgfz
yygfy
xxgfx
Chapter 5 Partial differentiation
)0,1,0()001( ,)0,2/1,2/1( :0
)0,0,1( ,)1,0,0( ,)2/1,0,2/1( :0for points stationary
0 and 0 case thefor ,Similarily
)0,1,0( ,)1,0,0( ),2/1,2/1,0( :points stationary
11 ,0230 ,0 (c)
11 ,0230 ,0 (b) 22
22
, ,,z
y
zy
yyλyyyzx
zzλzzzyx
6/2 ,6/1 ,6/1161 from
20 from , if (a)
0)(2)(30)(2)(3)2()1(
(3)---023)(
(2)---023)(
(1)---023)( (ii)
2222
22
2
2
2
zyxxzyx
xzzyxyx
yxyxyxyx
zzhgfz
yyhgfy
xxhgfx
Chapter 5 Partial differentiation
ntinconsistedifferent all are and , (d)
sconstraint by prohibited (c)
(a) condition in included
02)(3
02)(3
02)(3 if (b)
6/1 ,6/1 ,6/2 if
61 ,6/2 ,6/1 if ,similarily
zyx
zyx
zxzx
zyzy
yxyx
zyxzy
/zyxzx
Chapter 5 Partial differentiation
5.11 Thermodynamic relations
Maxwell’s thermodynamic relations:
P: pressure V: volume T: temperature S: entropy U: internal energy
XY
U
YX
UdY
Y
UdX
X
UdU
PdVTdSdU
XY
22
and )()(
:Y and Xof variables twofor
micsthermodyna of lawfirst the
Ex: Show that VS SPVT )/()/(
VS
SV
SV
S
P
V
T
VS
U
SV
U
PV
UT
S
U
dVV
UdS
S
UdUPdVTdS
)()(
)( and )(
)()(
22
Chapter 5 Partial differentiation
Ex: Show that VT TPVS )/()/(
VT
TVVT
VT
VVTT
VT
VT
VT
T
P
V
STV
ST
T
ST
VT
P
VT
ST
V
S
T
U
VV
U
TTV
U
VT
U
T
ST
T
UP
V
ST
V
U
dTT
UdV
V
U
PdVdTT
SdV
V
STPdVTdSdU
dTT
SdV
V
SdS
TVdUdS
)()(
])([)()()(
)()( since
)() ( and )()(
)()(
])()[(
)()(
and variablest independen two with and aldifferenti totalconsider
22
22
Chapter 5 Partial differentiation
5.12 Differentiation of integrals
dtx
txf
x
txFdttxf
xdt
x
txF
t
x
txf
t
txF
xx
txF
ttx
txF
xt
txF
txfx
txFdttxftxF
),(),(),(]
),([
),(]
),([]
),([
),(),(
),(),(
),(),(
22
indefinite integral
dtx
txf
dx
dI
dtx
txfdt
x
txf
x
uxF
x
vxF
dx
xdI
uxFvxFdttxfxI
v
u
uv
vt
ut
),(
),(),(),(),()(
),(),(),()(
definite integral
(1) The integral’s limits are constant:
Chapter 5 Partial differentiation
(2) The integral’s limits are function of x
dt x
txf
dx
duxx,uf
dx
dvxvxf
dx
dI
dttxfxdx
duxuxf
dx
dvxvxf
x
I
dx
du
u
I
dx
dv
v
I
dx
dI
xuxfu
Ixvxf
v
I
xuxFxvxFdttxfI
xv
xu
xv
xu
xvt
xut
)(
)(
)(
)(
)(
)(
),())((-))(,(
),())(,())(,(
))(,( ,))(,(
))(,())(.(),(
Ex: Find the derivative with respect to x of the integral
dtt
xtxI
x
x2 sin
)(
)sin2sin3(1
|sinsinsin2
cos)1(
sin)2(
sin
2323
2
2
3
2
2
xxxx
xt
x
x
x
x
dtt
xtt
x
xx
x
x
dx
dI
xx
x
x