2.1.1 partial derivatives of first order · x . alone. the derivative of . z. with respect to . x...
TRANSCRIPT
74 Engineering Mathematics
2.1.1 Partial Derivatives of First Order
Consider a function z = f (x, y) of two independent variables x and y. By keeping y as a constant and varying x only, z becomes a function of x alone. The derivative of z with respect to x (y is kept constant) is called the partial derivative of z with respect to x and is denoted by ∂ ∂ ∂ ∂z x f x f D f/ or / or orx x . Thus,
δ
δ∂∂
= + −δ →
zx
f x x y f x yx
lim ( , ) ( , )x 0
Similarly, the derivative of z with respect to y (when x is kept constant) is called the partial derivative of
z with respect to y and is denoted by ∂ ∂ ∂ ∂z y f y f D f/ or / or ory y . Thus,
δδ
∂∂
= + −δ →
zy
f x y y f x yy
lim ( , ) ( , )y 0
z/x and z/y are called the first order partial derivatives of z.
In general, if z is a function of more than two independent variables, then the partial derivative of z with respect to any one of the variables, keeping all other variables constant, is the partial derivative of z with respect to that variable.
2.1.2 Partial Derivatives of Higher Order
The first order partial derivatives z/x and z/y being the functions of x and y can be further differentiated partially with respect to x and y to get the second order partial derivatives. The second order partial derivatives are
∂∂
∂∂
=
∂∂x
zx
zx
for xx
2
2, ∂∂
∂∂
=
∂∂y
zy
zy
for yy
2
2, ∂∂
∂∂
=
∂∂ ∂y
zx
zy x
for xy
2, ∂∂
∂∂
=
∂∂ ∂x
zy
zx y
for yx
2
The derivatives fxy and fyx are called mixed derivatives. In general, 2z/yx = 2z/xy or fxy = fyx if fxy and fyx are continuous, i.e., the order of differentiation is immaterial if the partial derivatives involved are continuous.
In general, the first order partial derivatives z/x and z/y can be differentiated successively to the
partial derivatives of higher order.
Note: (i) If z = u + v, where u = f (x, y), v = f (x, y), then z is a function of x and y.
∂∂
= ∂∂
+ ∂∂
zx
ux
vx
, ∂∂
= ∂∂
+ ∂∂
zy
uy
vy
(ii) If z = uv, where u = f (x, y), v = f (x, y), then ∂∂zx
= ∂∂
= ∂∂
+ ∂∂
uvx
u vxv ux
( ) , ∂∂
= ∂∂
= ∂∂
+ ∂∂
zy
uvy
u vyv uy
( ) .
Functions of Several Variables 75
(iii) If z = u / v, where u = f (x, y), v = f (x, y), then ∂∂zx
= ∂∂
=
∂∂
− ∂∂u v
x
v uxu vx
v( / )
2, ∂∂
= ∂∂
=
∂∂
− ∂∂z
yu vy
v uyu vy
v( / )
2.
(iv) If z = f (u), where u = f (x, y), then ∂∂
= ⋅ ∂∂
zx
dzdu
ux
, ∂∂
= ⋅ ∂∂
zy
dzdu
uy
.
Example 2.1 Find all the partial derivatives of ax2 + 2hxy + by.
Solution: Let f (x, y) = ax2 + 2hxy + by2
Then ∂∂
= + ∂∂
= +fx
ax hy fy
hx by2 2 , 2 2
∂∂
= ∂∂
∂∂
=
∂∂
+ =fx x
fx x
ax hy a(2 2 ) 22
2
∂∂ ∂
= ∂∂
∂∂
=
∂∂
+ =fy x y
fx y
ax hy h(2 2 ) 22
∂∂ ∂
= ∂∂
∂∂
=
∂∂
+ =fx y x
fy x
hx by h(2 2 ) 22
and ∂∂
= ∂∂
∂∂
=
∂∂
+ =fy y
fy y
hx by b(2 2 ) 22
2
All the partial derivatives of order higher than two vanish. It may be observed that
∂∂ ∂
= ∂∂ ∂
fx y
fy x
2 2
Example 2.2 If z = f (x + ct) + j(x – ct), prove that ∂∂
= ∂∂
zt
c zx
2
22
2
2.
Solution: We have ϕ ϕ∂∂
= ′ + ⋅ ∂∂
+ + ′ − ∂∂
− = ′ + + ′ −zx
f x ctxx ct x ct
xx ct f x ct x ct( ) ( ) ( ) ( ) ( ) ( )
and ϕ∂∂
= ′′ + + ′′ −zx
f x ct x ct( ) ( )2
2 (1)
Again ϕ ϕ∂∂
= ′ + ∂∂
+ + ′ − ∂∂
− = ′ + − ′ −zt
f x cttx ct x ct
tx ct cf x ct c x ct( ) ( ) ( ) ( ) ( ) ( )
and ϕ ϕ∂∂
= ′′ + + ′′ − = ′′ + + ′′ − zt
c f x ct c x ct c f x ct x ct( ) ( ) ( ) ( )2
22 2 2 (2)
From (1) and (2), we get ∂∂
= ∂∂
zt
c zx
2
22
2
2
76 Engineering Mathematics
Example 2.3 If v = (x2 + y2 + z2)–1/2, prove that ∂∂
+ ∂∂
+ ∂∂
=vx
vy
vz
02
2
2
2
2
2.
Solution: We have ∂∂
= − + + −vx
x y z12
( )2 2 2 3/2 = − + + −x x x y z2 ( )2 2 2 3/2
and ∂∂
= − ⋅ + + + −
+ + ⋅− −v
xx y z x x y z x[1 ( ) 3
2( ) 2 ]
2
22 2 2 3/2 2 2 2 5/2
= − + + + + − = + + − −− −x y z x y z x x y z x y z( ) ( 3 ) ( ) (2 )2 2 2 5/2 2 2 2 2 2 2 2 5/2 2 2 2
Similarly, ∂∂
= + + − + −−vy
x y z x y z( ) ( 2 )2
22 2 2 5/2 2 2 2 and
∂∂
= + + − − +−vz
x y z x y z( ) ( 2 )2
22 2 2 5/2 2 2 2
Hence, ∂∂
+ ∂∂
+ ∂∂
= + + ⋅ =−vx
vy
vz
x y z( ) (0) 02
2
2
2
2
22 2 2 5/2
Note: The given equation is known as the Laplace equation, and the function which satisfies this equation is called the harmonic function.
Example 2.4 If θ =−
t enrt4
2
, find the value of n which will make θ θ∂
∂∂∂
=
∂∂r r
rr t
12
2 .
Solution: Let θ =−
t enrt4
2
θ∂∂
= ⋅ ⋅ −
= −
− − −
rt e r
trt e2
412
nrt n
rt4 1 4
2 2
\ θ∂∂
= − ⋅ − −r
rr t e1
2n
rt2 3 1 4
2
θ∂
∂∂∂
= − + −
= − −
− − − − −
rr
rt r e r e r
tt r e r
t12
3 24
12
32
nrt
rt n
rt2 1 2 4 3 4 1 2 4
22 2 2
\ θ∂
∂∂∂
= −
− −
r rr
rt e r
t1 1
2 23n
rt
22 1 4
22
Also, θ∂∂
= + ⋅
= +
− − − − −
tnt e t e r
tt e n r
t4 4n
rt n
rt n
rt1 4 4
2
21 4
22 2 2
Since θ θ∂
∂∂∂
=
∂∂r r
rt t
12
2 [Given]
or −
= +
− − − −t e r
tt e n r
t12 2
34
nrt n
rt1 4
21 4
22 2
Functions of Several Variables 77
or − = +rt
n rt4
32 4
2 2
or = −n 32
Example 2.5 If u = (1 – 2xy + y2)–1/2, prove that ∂∂
− ∂∂
+ ∂∂
∂∂
=
xx u
x yy u
y(1 ) 02 2 .
Solution: u = (1 – 2xy + y2)–1/2 = V–1/2, where V = 1 – 2xy + y2
∂∂
= − ⋅ ∂∂
= − − =− − −ux
V Vx
V y yV12
12
( 2 )3/2 3/2 3/2
∂∂
= ⋅ ∂∂
= ⋅ −
⋅ ∂
∂= − − =− − − −u
xyxV y V V
xyV y y V( ) 3
232
( 2 ) 32
23/2 5/2 5/2 2 5/2
\ ∂∂
− ∂∂
= − ⋅ ∂
∂+ ∂∂
⋅ ∂∂
−x
x ux
x ux
ux x
x(1 ) (1 ) (1 )2 22
22
= − ⋅ + − = − −− − − −x y V yV x yV yV x x(1 ) 3 ( 2 ) [3 (1 ) 2 ]2 2 5/2 3/2 3/2 1 2 (1)
Similarly, ∂∂
= − ∂∂
= − − + = ⋅ −− − −uy
V Vy
V x y V x y12
12
( 2 2 ) ( )3/2 3/2 3/2
∂∂
= ⋅ ∂∂
− + − ⋅ ∂∂
− −uy
Vyx y x y
yV( ) ( ) ( )
2
23/2 3/2
= ⋅ − + − ⋅ −
⋅∂∂
− −V x y V Vy
( 1) ( ) 32
3/2 5/2
= − − − ⋅ − + = − + −− − − −V x y V x y V x y V32
( ) ( 2 2 ) 3( )3/2 5/2 3/2 2 5/2
\ ∂∂
∂∂
= ⋅ ∂
∂+ ∂∂
⋅ ∂∂y
y uy
y uy
uy y
y( )2 22
22
= − + − + − ⋅− − −y V x y V V x y y[ 3( ) ] ( ) 22 3/2 2 5/2 3/2
= − + − + −− −yV y y x y V x y[ 3 ( ) 2( )]3/2 2 1
= − + −− −yV y x y V x y[3 ( ) (2 3 )]3/2 2 1 (2)
Adding (1) and (2), we have
∂∂
− ∂∂
+ ∂∂
∂∂
= − − + − + −− − −
xx u
x yy u
yyV yV x x y x y V x y(1 ) [3 (1 ) 2 3 ( ) 2 3 ]2 2 3/2 1 2 2 1
= − + − + −− −yV yV x x xy y y[3 (1 2 ) 3 ]3/2 1 2 2 2
= − + −− −yV yV xy y y[3 (1 2 ) 3 ]3/2 1 2
78 Engineering Mathematics
= −−yV y y(3 3 )3/2 = − +V xy y( 1 2 )2
= 0
Example 2.6 If +
++
++
=xa u
yb u
zc u
12
2
2
2
2
2 , prove that ∂∂
+ ∂
∂
+ ∂
∂
= ∂
∂+ ∂
∂+ ∂
∂
ux
uy
uz
x uxy uyz uz
22 2 2
Solution: We have x2(a2 + u)–1 + y2(b2 + u)–1 + z2(c2 + u)–1 = 1 (1)Differentiating (1) partially w.r.t. x, we get
+ − + ∂∂
− + ∂∂
− + ∂∂
=− − − −x a u x a u uxy b u u
xz c u u
x2 ( ) ( ) ( ) ( ) 02 1 2 2 2 2 2 2 2 2 2
or +
=+
++
++
∂∂
xa u
xa u
yb u
zc u
ux
2( ) ( ) ( )2
2
2 2
2
2 2
2
2 2
or ∂∂
=+
ux
xa u v
2( )2 , where =
++
++
+v x
a uy
b uz
c u( ) ( ) ( )
2
2 2
2
2 2
2
2 2
Similarly, differentiating (1) partially w.r.t. y, we get
+
=+
++
++
∂∂
yb u
xa u
yb u
zc u
uy
2( ) ( ) ( )2
2
2 2
2
2 2
2
2 2 or ∂∂
=+
uy
yb u v
2( )2
Similarly, differentiating (1) partially w.r.t. z, we get
+
=+
++
++
∂∂
zc u
xa u
yb u
zc u
uz
2( ) ( ) ( ) ( )2
2
2 2
2
2 2
2
2 2 or ∂∂
=+
uz
zc u v
2( )2
\ ∂∂
+ ∂
∂
+ ∂
∂
=
++
++
+
=u
xuy
uz v
xa u
yb u
zc u v
4( ) ( ) ( )
42 2 2
2
2
2 2
2
2 2
2
2 2 (2)
Also, ∂∂
+ ∂∂
+ ∂∂
=
++
++
+
x uxy uyz uz
xa u v
yb u v
zc u v
2 2 2( )
2( )
2( )
2
2
2
2
2
2
= +
++
++
=v
xa u
yb u
zc u v
4( ) ( ) ( )
42
2
2
2
2
2 [by (1)] (3)
From Eqs. (2) and (3), we get
∂∂
+ ∂
∂
+ ∂
∂
= ∂
∂+ ∂
∂+ ∂
∂
ux
uy
uz
x uxy uyz uz
22 2 2
Functions of Several Variables 79
Example 2.7 If u = log(x3 + y3 + z3 – 3xyz), show that
(i) ∂∂
+ ∂∂
+ ∂∂
= −
+ +x y zu
x y z9
( )
2
2
(ii) ∂∂
+ ∂∂
+ ∂∂
+ ∂∂ ∂
+ ∂∂ ∂
+ ∂∂ ∂
= −+ +
ux
uy
uz
uy z
uz x
ux y x y z
2 2 2 9( )
2
2
2
2
2
2
2 2 2
2
Solution: (i) u = log(x3 + y3 + z3 – 3xyz)
∂∂
= −+ + −
∂∂
= −+ + −
ux
x yzx y z xyz
uy
y zxx y z xyz
3 33
; 3 33
2
3 3 3
2
3 3 3
∂∂
= −+ + −
uz
z xyx y z xyz
3 33
2
3 3 3
Adding, ∂∂
+ ∂∂
+ ∂∂
= + + − − −+ + −
=+ +
ux
uy
uz
x y z xy yz zxx y z xyz x y z
3( )3
32 2 2
3 3 3
+ + − = + + + + − − − x y z xyz x y z x y z xy yz zx[ 3 ( )( )]3 3 3 2 2 2
Now, ∂∂
+ ∂∂
+ ∂∂
= ∂
∂+ ∂∂
+ ∂∂
∂∂
+ ∂∂
+ ∂∂
x y zu
x y z x y zu
2
= ∂∂
+ ∂∂
+ ∂∂
∂∂
+ ∂∂
+ ∂∂
=
∂∂
+ ∂∂
+ ∂∂
+ +
x y zux
uy
uz x y z x y z
3
= −+ +
−+ +
−+ +
= −+ +x y z x y z x y z x y z
3( )
3( )
3( )
9( )2 2 2 2 (1)
(ii) ∂∂
+ ∂∂
+ ∂∂
= ∂
∂+ ∂∂
+ ∂∂
∂∂
+ ∂∂
+ ∂∂
= ∂
∂+ ∂∂
+ ∂∂
∂∂
+ ∂∂
+ ∂∂
x y zu
x y z x y zu
x y zux
uy
uz
2
= ∂∂
∂∂
+ ∂∂
+ ∂∂
+
∂∂
∂∂
+ ∂∂
+ ∂∂
+
∂∂
∂∂
+ ∂∂
+ ∂∂
xux
uy
uz y
ux
uy
uz z
ux
uy
uz
= ∂∂
+ ∂∂ ∂
+ ∂∂ ∂
+ ∂∂ ∂
+ ∂∂
+ ∂∂ ∂
+ ∂∂ ∂
+ ∂∂ ∂
+ ∂∂
ux
ux y
ux z
uy x
uy
uy z
uz x
uz y
uz
2
2
2 2 2 2
2
2 2 2 2
2
= ∂∂
+ ∂∂
+ ∂∂
+ ∂∂ ∂
+ ∂∂ ∂
+ ∂∂ ∂
ux
uy
uz
uy x
uy z
uz x
2 2 22
2
2
2
2
2
2 2 2
∂∂ ∂
= ∂∂ ∂
∂∂ ∂
= ∂∂ ∂
∂∂ ∂
= ∂∂ ∂
uy x
ux y
uz y
uy z
ux z
uz x
, ,2 2 2 2 2 2
\ ∂∂
+ ∂∂
+ ∂∂
+ ∂∂ ∂
+ ∂∂ ∂
+ ∂∂ ∂
= ∂∂
+ ∂∂
+ ∂∂
= −
+ +ux
uy
uz
uy z
uz x
ux y x y z
ux y z
2 2 2 9( )
2
2
2
2
2
2
2 2 2 2
2 [From (1)]
80 Engineering Mathematics
2.2 HOMOGENEOUS FUNCTIONS AND EULER’S THEOREM
An expression of the form + + + +− −a x a x y a x y a yn n n
nn
0 11
22 2 in which each term is of nth degree is called
a homogeneous function of degree n in the variables x and y. It can be expressed as
+
+
+⋅ ⋅ ⋅+
x a a yx
a yx
a yx
nn
n
0 1 2
2
or
+
+
+⋅ ⋅ ⋅+
− −
y a xy
a xy
a xy
ann n n
n0 1
1
2
2
Hence, any function f (x, y) which can be expressed in the form φ φ
x y
xy x
yorn n is called a homogeneous
function of degree n in x and y. For example,
φ= ++
= ++
=
f x y x y
x yx y xx y x
x yx
( , ) (1 / )(1 / )
1/2
which means f (x, y) is a homogeneous function of degree 1/2 in x and y. Also,
φ= ++
= ++
=
f x y x y
x yy x yy x y
y xy
( , ) ( / 1)( / 1)
1/2
In general, a function f (x, y, t, …) is a homogeneous function of degree n in variables x, y, t, … if
φ= ⋅⋅ ⋅
f x y t x y
xtx
( , , , ) , ,n or φ ⋅⋅ ⋅
y x
yty
, ,n or φ ⋅⋅ ⋅
t x
tyt
, ,n , etc.
2.2.1 Euler’s Theorem on Homogeneous Functions
Theorem 2.1 If z is a homogeneous function of degree n in x and y, then ∂∂
+ ∂∂
=x zxy zynz.
Proof: Since z is a homogeneous function of degree n in x and y,
=
z x f yx
n
or ⇒ ∂∂
=
+ ′
⋅ ⋅ −
=
− ′
− − −z
xnx f y
xx f y
xyx
nx f yx
yx f yx
1n n n n12
1 2
or ∂∂
=
− ′
−x z
xnx f y
xyx f y
xn n 1 (2.9)
Also, ∂∂
= ′
⋅
= ′
−z
yx f y
x xx f y
x1n n 1 or
∂∂
= ′
−y z
yx yf y
xn 1 (2.10)
Adding (2.9) and (2.10), we get ∂∂
+ ∂∂
=
=x z
xy zynx f y
xnzn .
Note: In general, if z is a homogeneous function of degree n in x, y, t,..., then ∂∂
+ ∂∂
+ ∂∂
+⋅⋅ ⋅ =x zxy zyt zt
nz .
Functions of Several Variables 81
Deduction of Euler’s Theorem
Theorem 2.2 If z is a homogeneous function of degree n in x and y, then
∂∂
+ ∂∂ ∂
+ ∂∂
= −x zx
xy zx y
y zy
n n z2 ( 1)22
2
22
2
2
Proof: Since z is a homogeneous function of degree n in x and y, thus by Euler’s theorem, we have
∂∂
+ ∂∂
=x zxy zynz (2.11)
Differentiating (2.11) partially w.r.t. x, we get ∂∂
+ ∂∂
+ ∂∂ ∂
= ∂∂
zxx zx
y zx y
n zx
2
2
2 (2.12)
Differentiating (2.11) partially w.r.t. y, we get ∂∂
+ ∂∂ ∂
+ ∂∂
= ∂∂
zyx zy x
y zy
n zy
2 2
2
But ∂∂ ∂z
y x
2 = ∂
∂ ∂z
x y
2
Hence, ∂∂
+ ∂∂ ∂
+ ∂∂
= ∂∂
zyx zx y
y zy
n zy
2 2
2 (2.13)
Multiplying (2.12) by x, (2.13) by y, and then adding, we get
∂∂
+ ∂∂ ∂
+ ∂∂
+ ∂∂
+ ∂∂
= ∂∂
+ ∂∂
x z
xxy z
x yy z
yx zxy zyn x z
xy zy
222
2
22
2
2
or ∂∂
+ ∂∂ ∂
+ ∂∂
+ =x zx
xy zx y
y zy
nz n nz2 ( )22
2
22
2
2 [From (2.11)]
or ∂∂
+ ∂∂ ∂
+ ∂∂
= −x zx
xy zx y
y zy
n n z2 ( 1)22
2
22
2
2
Example 2.8 If = +−
−u x yx y
tan 13 3
, prove that ∂∂
+ ∂∂
=x uxy uy
usin 2 .
Solution: Here u is not a homogeneous function but = +−
=+
−=
u x y
x y
x y x
x y xx f y
xtan
1 ( / )
[1 ( / )]
3 3 3 32 is a
homogeneous function of degree 2 in x and y.
Therefore, by Euler’s theorem, we have
∂∂
+ ∂∂
=xx
u yy
u u(tan ) (tan ) 2 tan
or ∂∂
+ ∂∂
=x u uxy u u
yusec sec 2 tan2 2
or ∂∂
+ ∂∂
= ⋅ = =x uxy uy
uu
u u u u2sincos
cos 2sin cos sin 22
82 Engineering Mathematics
Example 2.9 If = ++
−u x yx y
cos 1 , show that ∂∂
+ ∂∂
+ =x uxy uy
u12
cot 0 .
Solution: Here, = ++
−u x yx y
cos 1 is not a homogeneous function but = ++
u x yx y
cos is homogeneous
in x and y of degree 1/2.Therefore, by Euler’s theorem, we have
∂∂
+ ∂∂
=xx
u yy
u ucos cos 12
cos
or − ∂∂
− ∂∂
=x u uxy u u
yusin sin 1
2cos
or ∂∂
+ ∂∂
+ =x uxy uy
u12
cot 0 . Hence the result.
Example 2.10 If = ++
−u x yx y
sin 1 , prove that
∂∂
+ ∂∂
=x uxy uy
u12
tan
and ∂∂
+ ∂∂ ∂
+ ∂∂
= −x ux
xy ux y
y uy
u uu
2 sin cos 24cos
22
2
22
2
2 3
Solution: Here u is not a homogeneous function but = = ++
z u x yx y
sin is a homogeneous function of
degree 1/2 in x and y. Therefore,
∂∂
+ ∂∂
=x zxy zy
z12
or ∂∂
+ ∂∂
=x u uxy u u
yucos cos 1
2sin
Thus, ∂∂
+ ∂∂
=x uxy uy
u12
tan (1)
Differentiating (1) partially w.r.t. x, we get
∂∂
+ ∂∂
+ ∂∂ ∂
= ∂∂
x ux
uxy ux y
u ux
12
sec2
2
22
or ∂∂
+ ∂∂ ∂
= −
∂∂
x ux
y ux y
u ux
12
sec 12
2
22 (2)
Again differentiating (1) partially w.r.t. y, we get
∂∂ ∂
+ ∂∂
+ ∂∂
= ∂∂
x uy x
y uy
uy
u uy
12
sec2 2
22
Functions of Several Variables 83
or ∂∂ ∂
+ ∂∂
= −
∂∂
x ux y
y uy
u uy
12
sec 12 2
22 (3)
Multiplying (2) by x and (3) by y and adding, we obtain
∂∂
+ ∂∂ ∂
+ ∂∂
= −
∂∂
+ ∂∂
x u
xxy u
x yy u
yu x u
xy uy
2 12
sec 122
2
22
2
22
or ∂∂
+ ∂∂ ∂
+ ∂∂
= −
x u
xxy u
x yy u
yu u2 1
2sec 1 1
2tan2
2
2
22
2
22 [by (1)]
= −uu
uu
14
sincos
12
sincos3
= − −u uu
sin (2cos 1)4cos
2
3
Hence, ∂∂
+ ∂∂ ∂
+ ∂∂
= −x ux
xy ux y
y uy
u uu
2 sin cos 24cos
22
2
22
2
2 3
2.3 TOTAL DERIVATIVE AND PARTIAL DIFFERENTIATION OF IMPLICIT FUNCTIONSIf φ ψ= = =z f x y x t y t( , ), where ( ), ( ) , then z is called a composite function of the single variable t. We can substitute the values of x and y in f(x, y) and express z as a function of t alone. We can then obtain dz/dt, which is called the total derivative. However, the substitution of φ ψ= =x t y t( ), ( ) is not always convenient. Then the use of partial derivatives helps in finding the total derivative. This method is given in the next section. Similarly, if φ ψ= = =z f x y x u v y u v( , ), where ( , ), ( , ) , then z is called a composite function of the two
variables u and v and we can find ∂∂
∂∂
zu
zv
and .
2.3.1 Chain Rule for Partial Differentiation
Theorem 2.3 If φ ψ= = =z f ( x y ) where x ( t ) y ( t ), , , i.e., z is a composite function of t, then
= ∂∂
⋅ + ∂∂
⋅dzdt
zxdxdt
zydydt (2.14)
Proof: We have φ ψ= = =z f x y x t y t( , ), where ( ), ( ) (2.15)
Let an increment to t be δt and the corresponding increments to x, y and z be δx , δ y and δz , respectively. Then δ δ δ+ = + +z z f x x y y( , ) (2.16)
84 Engineering Mathematics
Subtracting (2.15) from (2.16), we get δ δ δ= + + −z f x x y y f x y( , ) ( , )
δ δ δ δ= + + − + + + −f x x y y f x y y f x y y f x y{ ( , ) ( , )} { ( , ) ( , )}
Hence, δδ
δ δ δδ
δδ
δδ
δδ
= + + − + ⋅ + + − ⋅zt
f x x y y f x y yx
xt
f x y y f x yy
yt
{ ( , ) ( , )} { ( , ) ( , )}
Taking limit as δ →t 0 , δx and δ →y also 0 , we have
δδ
δ δ δδ
δδ
δδ
δδ
= + + − +
+ + −
δ δ
δδ δ δ→ →
→→ → →
zt
f x x y y f x y yx
xt
f x y y f x yy
yt
lim lim ( , ) ( , ) lim lim ( , ) ( , ) limt x
yt y t0 0
00 0 0
or dzdt
δ= ∂ +∂
⋅ + ∂
∂⋅
δ →
f x y yx
dxdt
f x yy
dydt
lim ( , ) ( , )y 0
[Assuming ∂ ∂f x y x( , ) / as a continuous function of y]
= ∂∂
⋅ + ∂∂
⋅f x yx
dxdt
f x yy
dydt
( , ) ( , ) = ∂∂
⋅ + ∂∂
⋅zxdxdt
zydydt
which is known as the chain rule.
Note:
(i) Taking t = x, (2.14) becomes = ∂∂
+ ∂∂
⋅dzdx
zx
zydydx
(2.17)
(ii) If z = f (x, y, s,...), where x, y, s, … are all functions of a variable t, then
= ∂∂
⋅ + ∂∂
⋅ + ∂∂
⋅ + ⋅ ⋅ ⋅dzdt
zxdxdt
zydydt
zsdsdt
(iii) If =z f x y( , ) and x, y are functions of u and v, then
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
zu
zx
xu
zy
yu
zv
zx
xv
zy
yv
,
2.3.2 Differentiation of Implicit Functions
If =f x y( , ) constant, c (say) is an implicit relation between x and y, i.e., y is an implicit function of x, then
=dfdx
0 or ∂∂
+ ∂∂
⋅ =fx
fydydx
0 [Using (2.17)]
This implies that = − ∂ ∂∂ ∂
= −dydx
f xf y
ff
//
x
y
∂∂
≠
fy
0 (2.18)
which is the first differential coefficient of an implicit function.
We can also express d ydx
2
2 in terms of partial derivatives. Let
∂∂
= ∂∂
= ∂∂
= ∂∂
= ∂∂ ∂
=fx
p fy
q fx
r fy
t fx y
s, , , ,2
2
2
2
2
.
Functions of Several Variables 85
= −dydx
pq
[From (2.18)]
On differentiating again w.r.t. x, we obtain
= − −d ydx
q dp dx p dq dxq
( ) ( )2
2 2. (2.19)
But by chain rule, = ∂∂
⋅ + ∂∂
⋅ = + −
=
−dpdx
px
pydydx
r s pq
qr psq
1
and = ∂∂
⋅ + ∂∂
⋅ = + −
=
−dqdx
qx
qydydx
s t pq
qs ptq
1
Substituting the values of dp dx and dq dx in (2.19), we get
= − −
−
−
= − − +d y
dx qq qr ps
qp qs pt
q qq r pqs p t1 1 ( 2 )
2
2 2 32 2
\ = − − +d ydx
q r pqs p tq
22
2
2 2
3
which is the second differential coefficient of an implicit function.
Example 2.11 If =+ +
+ + ++ +
u x y z
x y zxy yz zxx y z
log3 3 3
3 3 3 2 2 2 , then find ∂∂
+ ∂∂
+ ∂∂
x uxy uyz uz
.
Solution: Let =+ +
v x y zx y z
3 3 3
3 3 3 and = + ++ +
w xy yz zx
x y zlog
2 2 2 . (1)
so that u = v + w
Since =+
v x y x z xy x z x
( / ) ( / )1 ( / ) ( / )
63 3
3 3, v is a homogeneous function of degree 6 in x, y, z.
Hence, by Euler’s theorem, we have
∂∂
+ ∂∂
+ ∂∂
=x vxy vyz vz
v6 (2)
Since =+ ⋅ +
+
+
w
yx
yxzxzx
yx
zx
log
12 2 , therefore w is a homogeneous function of degree zero in x, y, z.
Hence, by Euler’s theorem, we have
∂∂
+ ∂∂
+ ∂∂
=x wxy wyz wz
0 (3)
86 Engineering Mathematics
Adding (2) and (3), we obtain
∂∂
+ ∂∂
+
∂∂
+ ∂∂
+
∂∂
+ ∂∂
=x v
xwx
y vy
wy
z vz
wz
v6
or ∂∂
+ ∂∂
+ ∂∂
=+ +
x uxy uyz uz
x y zx y z
63 3 3
3 3 3
Example 2.12 If z is a function of x and y, where = + −x e eu v and = −−y e eu v , show that
∂∂
− ∂∂
= ∂∂
− ∂∂
zu
zv
x zxy zy
Solution: Here z is a composite function of u and v. Therefore,
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
= ∂∂
⋅ + ∂∂
− −zu
zx
xu
zy
yu
zxe z
ye( )u u
and ∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
= ∂∂
− + ∂∂
−−zv
zx
xv
zy
yv
zx
e zy
e( ) ( )v v
Subtracting, ∂∂
− ∂∂
= + ∂∂
− − ∂∂
= ∂∂
− ∂∂
− −zu
zv
e e zx
e e zyx zxy zy
( ) ( )u v u v
Example 2.13 (i) If + = + +y x x y( )x y x y( ) , find dydx
.
(ii) If = +z x y2 2 and + + =x y axy a3 53 3 3 , find the value of dzdx
when x = y = a.
Solution: (i) Let = + − + +f x y y x x y( , ) ( )x y x y( )
Now, ∂∂
= + ⋅ − + + +− +fx
y y y x x y x ylog ( ) [1 log( )]x y x y1
and ∂∂
= + − + + +− +fy
x y x x x y x ylog ( ) [1 log( )]x y x y1
Hence, ( )( )
= − ∂ ∂∂ ∂
= −+ − + + +
+ − + + +
− +
− +dydx
f xf y
y y y x x y x y
x y x x x y x y//
log ( ) 1 log( )
log ( ) 1 log( )
x y x y
x y x y
1
1
(ii) Since = +z x y2 2 , thus by the concept of total derivative, we have
= ∂∂
+ ∂∂
dz zxdx z
ydy
= ∂∂
⋅ + ∂∂
dzdx
zx
zydydx
1
= + ⋅ + + ⋅ ⋅ =+
+
− −x y x x y y dy
dx x yx y dy
dx12
( ) 2 1 12
( ) 2 12 2 1/2 2 2 1/2
2 2 (1)
Functions of Several Variables 87
From the relation + + =x y axy a3 53 3 3 , we have
=− ∂∂
+ + −
∂∂
+ + −= − +
+= −dy
dxxx y axy a
yx y axy a
x ayy ax
( 3 5 )
( 3 5 )
(3 3 )3 3
13 3 3
3 3 3
2
2 at (a, a)
Putting = −dydx
1 in Eq. (1), we get
=
++ −
=dzdx x y
x y1 { ( 1)} 0a a a a( , ) 2 2
( , )
Example 2.14 If θ θ= = =w f x y x r y r( , ), cos , sin , show that
θ
∂∂
+ ∂
∂
= ∂
∂
+ ∂
∂
wr r
w fx
fy
12
2
2 2 2
Solution: The given equation defines w as a composite function of r and q.
θ θ∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
= ∂∂
⋅ + ∂∂
⋅wr
wx
xr
wy
yr
wx
wy
cos sin
or θ θ∂∂
= ∂∂
+ ∂∂
wr
fx
fy
cos sin =w f x y[ ( , )] (1)
Also, θ θ θ
θ θ∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
= ∂∂
− + ∂∂
w wx
x wy
y wx
r wyr( sin ) ( cos )
or θ
θ θ∂∂
= −∂∂
+ ∂∂r
w fx
fy
1 sin cos (2)
Squaring and adding (1) and (2), we get
θ
∂∂
+ ∂
∂
= ∂
∂
+ ∂
∂
wr r
w fx
fy
12
2
2 2 2
Example 2.15 If = − −
u u y x
xyz xxz
, , show that ∂∂
+ ∂∂
+ ∂∂
=x uxy u
yz u
z02 2 2 .
Solution: Let = − = −v y xxy x y
1 1 and = − = −w z xxz x z
1 1 (1)
so that =u u v w( , ) . Therefore,
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
= ∂∂
−
+
∂∂
−
ux
uv
vx
uw
wx
uv x
uw x
1 12 2 [using (1)]
88 Engineering Mathematics
or ∂∂
= −∂∂
− ∂∂
x ux
uv
uw
2 (2)
Also, ∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
= ∂∂
+
∂∂
uy
uv
vy
uw
wy
uv y
uw
1 (0)2
[using (1)]
or ∂∂
= ∂∂
y uy
uv
2 (3)
Similarly, ∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
= ∂∂
+ ∂∂
uz
uv
vz
uw
wz
uv
uw z
(0) 12
[using (1)]
or ∂∂
= ∂∂
z uz
uw
2 (4)
Adding (2), (3) and (4), we have
∂∂
+ ∂∂
+ ∂∂
=x uxy u
yz u
z02 2 2
Example 2.16 If u x y f yx
( )= −
find x
ux
xy ux y
yuy
222
2
22
2
2∂∂
+ ∂∂ ∂
+ ∂∂
Solution: Given : ux
x y fyx
yx
f yx
( ) '2
∂∂
= −
−
+
yx
x y fyx
f yx
( )2
'= −
−
+
ux
yx
x y fyx
yx
yx
fyx
yx
x y fyx
fyx
( ) 2 ( )yx
2
2 2''
2 2'
3' '
2∂∂
= −
−
−
+
−
+
−
+
−
x
ux
yx
x y fyx
y fyx
yx
x y fyx
y f( ) 2 ( )yx2
2
2
2'' ' ' '∂
∂= −
−
−
+
−
−
x
ux
yx
x y fyx
y fyx
yx
x y fyx
( ) 2 2 ( )22
2 2'' ' '∂
∂=
−
−
+
−
∂∂
= −
+
−u
yx y f
yx x
f yx
( ) 1 ( 1)'
= −
−
xx y f
yx
f yx
1 ( ) '
∂∂
=
−
+
− −
uy x
x y fyx x x
fyx
fyx x
1 ( ) 1 1 ( 1) 12
2'' ' '
=
−
−
xx y f
yx x
fyx
1 ( ) 22
'' '
Functions of Several Variables 89
∂∂
=
−
−
y
uy
yx
x y fyx
yxfyx
( )22
2
2
2
2''
2'
∂∂ ∂
=
−
−
+
+
−
−
−
−
u
x y xx y f
yx
yx x
fyx x
x y fyx
fyx
1 ( ) 1 1 ( )yx
2''
2'
2' '
2
= −
−
+
+
−
−
−
−
y
xx y f
yx x
fyx x
x y fyx
fyx
( ) 1 1 ( )yx
3' '
2' '
2
∂∂ ∂
=−
−
+
−
−
+
xy u
x yyx
x y fyx
y fyx
y x yx
fyx
yxf2
2( ) 2 2 ( ) 2 y
x
2 2
3' ' '
2'
∂∂
+ ∂∂ ∂
+ ∂∂
=xux
xy ux y
yuy
2 022
2
22
2
2
Example 2.17 If W=f(y z, z x, x y) show that ∂∂
+ ∂∂
+ ∂∂
=Wx
Wy
Wz
0
Solution: Let u = y z, v = z x, w = x y, W = f (u, v, w)
∂∂
= ∂∂
∂∂
+ ∂∂
∂∂
+ ∂∂
∂∂
= ∂∂
+ ∂∂
− + ∂∂
= −∂∂
+ ∂∂
Wx
fuux
fvvx
fwwx
fu
fv
fw
fv
fw
(0) ( 1) (1)
∂∂
= ∂∂
∂∂
+ ∂∂
∂∂
+ ∂∂
∂∂
= ∂∂
+ ∂∂
+ ∂∂
− = ∂∂
− ∂∂
Wy
fuuy
fvvy
fwwy
fu
fv
fw
fu
fw
(1) (0) ( 1)
∂∂
= ∂∂
∂∂
∂∂
∂∂
+ ∂∂
∂∂
= ∂∂
− + ∂∂
+ ∂∂
= −∂∂
+ ∂∂
Wz
fuuzfvvz
fwwz
fu
fv
fw
fu
fv
( 1) (1) (0)
Adding we get ∂∂
+ ∂∂
+ ∂∂
=Wx
Wy
Wz
0
2.4 CHANGE OF VARIABLESLet =z f x y( , ) (2.20) where φ ψ= =x u v y u v( , ) and ( , ) (2.21) More often, it is required to change expressions containing z, x, y, ∂ ∂z x/ , ∂ ∂z y/ , etc., to expressions containing z, u, v, ∂ ∂z u/ , z v/∂ ∂ , etc. Then we can obtain the necessary formulae for the change of variables. The variables x, y, z will be functions of u alone when v is treated as a constant. Thus, by chain rule, we have
zu
zx
xu
zy
yu
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
(2.22)
where the ordinary derivatives are being changed to the partial derivatives since x, y are functions of two
variables u and v. Similarly, treating u as a constant, we obtain zv
zx
xv
zy
yv
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
(2.23)
90 Engineering Mathematics
Solving (2.22) and (2.23) for zx∂∂
and zy∂∂
, we get the values in terms of zu∂∂
, zv∂∂
, z, u, v.
Note: (i) If instead of Eq. (2.21), we are given that u x y v x y( , ) and ( , )φ ψ= = , then (2.22) and (2.23) become
zx
zu
ux
zv
vx
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
and zy
zu
uy
zv
vy
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
(ii) By repeatedly applying formulae (2.22) and (2.23) or the formulas in Note (i), we can obtain the higher derivatives of z.
Example 2.18 If u F x y y z z x( , , )= − − − , prove that ux
uy
uz
0∂∂
+ ∂∂
+ ∂∂
=
Solution: Put x y r y z s, ,− = − = and z x t− = , so that u f r s t( , , )= . Therefore,
ux
ur
rx
us
sx
ut
tx
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
ur
xs
ut
ur
ut
(1) (0) ( 1)= ∂∂
⋅ + ∂∂
⋅ + ∂∂
⋅ − = ∂∂
− ∂∂
(1)
Similarly, uy
ur
ry
us
sy
ut
ty
ur
us
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
= − ∂∂
+ ∂∂
(2)
and uz
ur
rz
us
sz
ut
tz
us
ut
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
= − ∂∂
+ ∂∂
(3)
Adding (1), (2) and (3), we get ux
uy
uz
0∂∂
+ ∂∂
+ ∂∂
=
Example 2.19 If z f x y( , )= and x r cosθ= , y r sinθ= show that
zx
zy
zr r
z12 2 2
2
2
θ∂∂
+ ∂
∂
= ∂
∂
+ ∂
∂
Solution: Here, zr
zx
xr
zy
yr
zx
zy
cos sinθ θ∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
= ∂∂
+ ∂∂
(1)
and z zx
x zy
y zx
r zyr( sin ) ( cos )
θ θ θθ θ∂
∂= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
= ∂∂
− + ∂∂
rz z
xzy
1 sin cosθ
θ θ∂∂
= − ∂∂
+ ∂∂
(2)
Functions of Several Variables 91
Squaring and adding (1) and (2), we get
zr r
z zx
zy
zxzy
1 cos sin 2sin cos2
2
2 22
22
θθ θ θ θ∂
∂
+ ∂
∂
= ∂
∂
+ ∂
∂
+ ∂
∂∂∂
zx
zy
zxzy
sin cos 2sin cos2
22
2θ θ θ θ+ ∂∂
+ ∂
∂
− ∂
∂∂∂
zx
zy
2 2
= ∂∂
+ ∂
∂
Example 2.20 Transform the equation ux
uy
02
2
2
2∂∂
+ ∂∂
= into polar coordinates.
Solution: The relation which connects Cartesian coordinates (x, y) with polar coordinates (r, q) are x r cosθ= ,
y r sinθ= so that r x y2 2 2= + and yx
tanθ = . Thus,
r x y yx
and tan2 2 1θ= + =
−
\ rx
x
x y
rr
cos cos2 2
θ θ∂∂
=+
= = , ry
y
x y
rr
sin sin2 2
θ θ∂∂
=+
= =
and x y
x
yx
yx y
rr r
1
1
sin sin2
2
2 2 2 2θ θ θ∂∂
=+
⋅ −
= −
+= − = − ;
y yx
xx
x yrr r
1
1
1 cos cos2
2
2 2 2θ θ θ∂∂
=+
⋅ =+
= =
Here, u is a composite function of x and y.
ux
ur
rx
ux
ur r
ucos sinθ
θ θ θθ
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
= ∂∂
− ⋅ ∂∂
xu
r ru( ) cos sinθ θ
θ∂∂
= ∂∂
− ⋅ ∂∂
or x r r
cos sinθ θθ
∂∂
≡ ∂∂
− ⋅ ∂∂
(1)
Also, uy
ur
ry
uy
ur r
usin cosθ
θ θ θθ
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
= ∂∂
+ ⋅ ∂∂
or yu
r ru( ) sin cosθ θ
θ∂∂
= ∂∂
+ ⋅ ∂∂
or
y r rsin cosθ θ
θ∂∂
≡ ∂∂
+ ⋅ ∂∂
(2)
92 Engineering Mathematics
Now, we will make use of the equivalence of Cartesian and polar operators as given by (1) and (2).
ux x
ux r r
ur r
ucos sin . cos sin2
2θ θ
θθ θ
θ∂∂
= ∂∂
∂∂
=
∂∂
− ∂∂
∂∂
− ⋅ ∂∂
r
ur r
ur
ur r
ucos cos sin sin cos sinθ θ θθ
θθ
θ θθ
= ∂∂
∂∂
− ∂∂
− ⋅ ∂
∂∂∂
− ⋅ ∂∂
ur
ur r
ur
cos cos sin 1 sin2
2 2
2θ θ θ
θθ
θ= ∂
∂− ∂
∂−
− ⋅ ∂
∂ ∂
r
ur
ur r
ur
usin sin cos cos sin2 2
2θ θ θ
θθ
θθ
θ− − ∂
∂+ ∂
∂ ∂− ∂
∂− ⋅ ∂
∂
ur r
ur
ur r
ur r
ucos 2cos sin sin 2cos sin sin22
2 2
2 2 2
2
2
2θ θ θ
θθ θ θ
θθ
θ= ∂
∂+ ⋅ ∂
∂+ ⋅ ∂
∂− ⋅ ∂
∂ ∂+ ⋅ ∂
∂ (3)
uy y
uy r r
ur r
usin cos sin cos2
2θ θ
θθ θ
θ∂∂
= ∂∂
∂∂
=
∂∂
+ ⋅ ∂∂
∂∂
+ ⋅ ∂∂
r
ur r
ur
ur r
usin sin cos cos sin cosθ θ θθ
θθ
θ θθ
= ∂∂
∂∂
+ ⋅ ∂∂
+ ⋅ ∂
∂∂∂
+ ⋅ ∂∂
=ur r
ur
ur u r
ur
ur u r
ur
usin sin cos cos cos cos sin sin cos2
2 2
2 2 2
2θ θ θ
θθ θ θ θ θ
θθ
θ∂∂
− ⋅ ∂∂
+ ⋅ ∂∂ ⋅∂
+ ∂
∂+ ∂
∂ ⋅∂− ⋅ ∂
∂+ ⋅ ∂
∂
ur r
ur
ur r
ur r
usin 2cos sin cos 2cos sin cos22
2 2
2 2 2
2
2
2θ θ θ
θθ θ θ
θθ
θ= ∂
∂− ⋅ ∂
∂+ ⋅ ∂
∂+ ⋅ ∂
∂ ∂+ ⋅ ∂
∂ (4)
Adding (3) and (4), we get
ux
uy
ur r
ur r
u1 12
2
2
2
2
2 2
2
2θ∂∂
+ ∂∂
= ∂∂
+ ∂∂
+ ⋅ ∂∂
Therefore, ux
uy
02
2
2
2∂∂
+ ∂∂
= transforms into ur r
ur r
u1 1 02
2 2
2
2θ∂∂
+ ⋅ ∂∂
+ ⋅ ∂∂
= .
Example 2.21 If x y e2 cosφ+ = θ and x y ie2 sinφ− = θ , show that u u xy u
x y4
2
2
2
2
2
θ φ∂∂
+ ∂∂
= ∂∂ ∂
.
Solution: We have x e i e e(cos sin ) iφ φ= + = ⋅θ θ φ
and y e i e e(cos sin ) iφ φ= − = ⋅θ θ φ−
Functions of Several Variables 93
Here, u is a composite function of q and φ. Therefore,
u u
xx u
yy
θ θ θ∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
uxe e u
ye e x u
xy uy
( ) ( )i i= ∂∂
⋅ + ∂∂
⋅ = ∂∂
+ ∂∂
θ φ θ φ−
or xuyyθ
∂∂
= ∂∂
+ ∂∂
(1)
Also, u ux
x uy
yφ φ φ∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
uxie e u
ye ie ix u
xiy uy
( ) ( )i i= ∂∂
⋅ ⋅ + ∂∂
⋅ ⋅− = ∂∂
− ∂∂
θ φ θ φ−
or ixxiyyφ
∂∂
= ∂∂
− ∂∂
(2)
Using the operator (1), we have
u u xxyyx uxy uy
2
2θ θ θ∂∂
= ∂∂
∂∂
=
∂∂
+ ∂∂
∂∂
+ ∂∂
xxx ux
xxy uy
yyx ux
yyy uy
= ∂∂
∂∂
+
∂∂
∂∂
+
∂∂
∂∂
+
∂∂
∂∂
x x ux
ux
xy ux y
yx uy x
y y uy
uy
2
2
2 2 2
2= ∂
∂+ ∂∂
+
∂∂ ∂
+ ∂∂ ∂
+ ∂∂
+ ∂∂
x ux
xy ux y
y uy
x uxy uy
222
2
22
2
2= ∂
∂+ ∂
∂ ∂+ ∂
∂+ ∂
∂+ ∂
∂ (3)
Similarly, using operator (2), we get
u u ix
xiyyix uxiy uy
2
2φ φ φ∂∂
= ∂∂
∂∂
=
∂∂
− ∂∂
∂∂
− ∂∂
x ux
xy ux y
y uy
x uxy uy
222
2
22
2
2= − ∂
∂+ ∂
∂ ∂− ∂
∂− ∂
∂− ∂
∂ (4)
Adding (3) and (4), we get
u u xy u
x y4
2
2
2
2
2
θ φ∂∂
+ ∂∂
= ∂∂ ∂
94 Engineering Mathematics
2.5 JACOBIANS
Let u and v be functions of two independent variables x and y. Then the determinant u x u yv x v y
/ // /
∂ ∂ ∂ ∂∂ ∂ ∂ ∂
is called
the Jacobian of u and v with respect to x and y. It is denoted by u vx y
J u vx y
( , )( , )
or ,,
∂∂
.
Also, the Jacobian of u, v and w with respect to x, y and z is
u v wx y z
J u v wx y z
u x u y u zv x v y v zw x w y w z
( , , )( , , )
or , ,, ,
/ / // / // / /
∂∂
=
∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂
Similarly, we can define the Jacobian of more than three variables. The term ‘Jacobian’ was named after German mathematician Carl Gustav Jacob Jacobi (1804–1851) who made significant contribution to mechanics, partial differential equations, astronomy, elliptic functions and the calculus of variations.
2.5.1 Chain Rule for Jacobians
If u, v are functions of x, y and x, y are functions of r, s, then
u vx y
( , )( , )∂∂
. x yr s
( , )( , )
∂∂
= u vr s
( , )( , )∂∂
Let us prove this result.
u vx y
( , )( , )∂∂
. x yr s
( , )( , )
∂∂
= u x u yv x v y
/ // /
∂ ∂ ∂ ∂∂ ∂ ∂ ∂
x r x sy r y s
/ // /
∂ ∂ ∂ ∂∂ ∂ ∂ ∂
= u x u yv x v y
/ // /
∂ ∂ ∂ ∂∂ ∂ ∂ ∂
x r y rx s y s
/ // /
∂ ∂ ∂ ∂∂ ∂ ∂ ∂
(Interchanging rows and columns)
ux
xr
uy
yr
ux
xs
uy
ys
vx
xr
vy
yr
vx
xs
vy
ys
=
∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
Since u f x y( , )= , v g x y( , )= , where x r s( , )φ= , y r s( , )ψ= , thus by chain rule, we have
ur
ux
xr
uy
yr
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
, us
ux
xs
uy
ys
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
vr
vx
xr
vy
yr
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
, vs
vx
xs
vy
ys
∂∂
= ∂∂
⋅ ∂∂
+ ∂∂
⋅ ∂∂
Hence, u vx y
( , )( , )∂∂
. x yr s
( , )( , )
∂∂
= u r u sv r v s
/ // /
∂ ∂ ∂ ∂∂ ∂ ∂ ∂
= u vr s
( , )( , )∂∂
Functions of Several Variables 95
In general, x x xu u u
u u uv v v
x x xv v v
( , ,..., )( , ,..., )
( , ,..., )( , ,..., )
( , ,..., )( , ,..., )
n
n
n
n
n
n
1 2
1 2
1 2
1 2
1 2
1 2
∂∂
⋅∂∂
=∂∂
Corollary: If J =u vx y
( , )( , )∂∂
and Jx yu v
( , )( , )
′ = ∂∂
, then JJ 1′ = .
Proof: If we replace r, s by u, v in the previous result, then we get the required result. In general,
u u ux x x
x x xu u u
( , ,..., )( , ,..., )
( , ,..., )( , ,..., )
1n
n
n
n
1 2
1 2
1 2
1 2
∂∂
⋅∂∂
=
Example 2.22 If x r cosθ= , y r sinθ= , evaluate x yr
( , )( , )θ
∂∂
and rx y
( , )( , )
θ∂∂
.
Solution: Since x r cosθ= , y r sinθ= , we have
xr
cosθ∂∂
= , yr
sinθ∂∂
= and x r sinθ
θ∂∂
= − , y r cosθ
θ∂∂
=
x yr
xr
x
yr
yrr
( , )( , )
cos sinsin cosθ
θ
θ
θ θθ θ
∂∂
=
∂∂
∂∂
∂∂
∂∂
= −
r rcos sin2 2θ θ= + r r(cos sin )2 2θ θ= + =
Now, r x y2 2 2= + and yx
tan 1θ = −
rx
xr
∂∂
= , ry
yr
∂∂
= and x
yx y
yr2 2 2
θ∂∂
= −+
= − , y
xx y
xr2 2 2
θ∂∂
=+
=
rx y
rx
ry
x y
xr
yr
yr
xr
( , )( , )
2 2
θθ θ
∂∂
=
∂∂
∂∂
∂∂
∂∂
=−
xr
yr
x yr
rr r
12
3
2
3
2 2
3
2
3= + = + = =
Note: x yr
rx y
rr
( , )( , )
( , )( , )
1 1θ
θ∂∂
× ∂∂
= ⋅ =
Example 2.23 If x r sin cosθ φ= , y r sin sinθ φ= and z r cosθ= , show that x y zr
r( , , )( , , )
sin2
θ φθ∂
∂= .
Solution: We have x y zr
x r x xy r y yz r z z
r rr rr
( , , )( , , )
sin cos cos cos sin sinsin sin cos sin sin cos
cos sin 0θ φ
θ φθ φθ φ
θ φ θ φ θ φθ φ θ φ θ φ
θ θ
∂∂
=∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂
=−
−
96 Engineering Mathematics
Taking out common factors (r from the second column and r sin q from the third column), we get
x y zr
, ,, ,θ φ
∂( )∂( )
r sinsin cos cos cos sinsin sin cos sin cos
cos sin 0
2 θθ φ θ φ φθ φ θ φ φ
θ θ=
−
−
Expanding by third row, we get
x y zr
, ,, ,θ φ
∂( )∂( )
r sin coscos cos sincos sin cos
sinsin cos sinsin sin cos
2 θ θθ φ φθ φ φ
θθ φ φθ φ φ
=−
+−
= r2 sinq [cosq (cosq cos2f + cosq sin2f) + sinq (sinq cos2f + sinq sin2f)] r rsin (cos sin ) sin2 2 2 2θ θ θ θ= + =
Note: Here x y z( , , ) and θ φr( , , ) are the Cartesian and spherical polar coordinates of a point, respectively.
Example 2.24 If = = =yx xx
yx xx
yx xx
, ,12 3
12
3 1
23
1 2
3, show that the Jacobian of y y y, ,1 2 3 with respect to
x x x, ,1 2 3 is 4.
Solution: The Jacobian of y1, y2, y3 w.r.t. x1, x2, x3 is given by
∂∂
=
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
=
−
−
−
y y yx x x
yx
yx
yx
yx
yx
yx
yx
yx
yx
x xx
xx
xx
xx
x xx
xx
xx
xx
x xx
( , , )( , , )
1 2 3
1 2 3
1
1
1
2
1
3
2
1
2
2
2
3
3
1
3
2
3
3
2 3
12
3
1
2
1
3
2
3 1
22
1
1
2
3
1
3
1 2
32
Taking out common x x x1 , 1 and 1
12
22
32 from first, second and third row, respectively, we get
y y yx x x
( , , )( , , )
1 2 3
1 2 3
∂∂
=
−
−
−x x x
x x x x x x
x x x x x x
x x x x x x
1
12
22
32
2 3 3 1 1 2
2 3 3 1 1 2
2 3 3 1 1 2
Taking out common x x x x x x, and2 3 3 1 1 2 from the first, second and third column, respectively, we get
y y yx x x
( , , )( , , )
1 2 3
1 2 3
∂∂ =
−−
−
x x xx x x
1 1 11 1 11 1 1
12
22
33
12
22
33
= − − − − − + + = + + =1(1 1) 1( 1 1) 1(1 1) 0 2 2 4
Example 2.25 If = −u x y2 2, =v xy2 and θ=x r cos , θ=y r sin , find θ
∂∂u vr
( , )( , )
.
Solution: We have θ θ
∂∂
= ∂∂
× ∂∂
u vr
u vx y
x yr
( , )( , )
( , )( , )
( , )( , )
Functions of Several Variables 97
Since = − =u x y v xy, 22 2 , we have
∂∂
=
∂∂
∂∂
∂∂
∂∂
=−
= +u vx y
ux
uy
vx
vy
x yy x
x y( , )( , )
2 22 2
4( )2 2 (1)
Since θ θ= =x r y rcos , sin , we have
θθ
θ
θ θθ θ
∂∂
=
∂∂
∂∂
∂∂
∂∂
= − =x yr
xr
x
yr
yrr
r( , )( , )
cos sinsin cos
(2)
Hence, θ θ
θ θ∂∂
= ∂∂
⋅ ∂∂
= + ⋅ = + ⋅ =u vr
u vx y
x yr
x y r r r r r( , )( , )
( , )( , )
( , )( , )
4( ) 4( cos sin ) 42 2 2 2 2 2 3 [using (1) and (2)]
2.5.2 Jacobians of Implicit Functions
If u1 and u2 are implicit functions of the variables x1 and x2 connected by the relations =f u u x x( , , , ) 01 1 2 1 2
and =f u u x x( , , , ) 02 1 2 1 2 , then ∂∂
= −∂ ∂∂ ∂
u ux x
f f x xf f u u
( , )( , )
( 1)( , ) / ( , )( , ) / ( , )
1 2
1 2
2 1 2 1 2
1 2 1 2
. In general,
∂∂
= −∂ ∂∂ ∂
u u ux x x
f f f x x xf f f u u u
( , ,..., )( , ,..., )
( 1)( , ,..., ) / ( , ,..., )( , ,..., ) / ( , ,..., )
n
n
n n n
n n
1 2
1 2
1 2 1 2
1 2 1 2
Note: This result bears resemblance to the result ∂∂
= −∂ ∂∂ ∂
yx
f xf y
//
, where x and y are connected by the
relation =f x y( , ) 0 .
Example 2.26 If =u xyz, = + +v x y z2 2 2 and = + +w x y z , find ∂∂x y zu v w
( , , )( , , )
.
Solution: Let us first calculate the value of J.
= ∂∂
=
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
J u v wx y z
ux
uy
uz
vx
vy
vz
wx
wy
wz
( , , )( , , )
= =− −− −
yz zx xyx y z
yz z x y y x zx y x z x2 2 2
1 1 1
( ) ( )2 2( ) 2( )1 0 0
= − − − − − = − − −z x y z x y y x x z x y x z y z2 ( )( ) 2 ( )( ) 2( )( )( )
Since ′ =JJ 1 , we have
∂∂
= ′ = =− − −
x y zu v w
JJ x y y z x z
( , , )( , , )
1 12( )( )( )
98 Engineering Mathematics
Example 2.27 If =−
=−
=−
u xy z
v yz x
w zx y
, , , show that ∂∂
=u v wx y z
( , , )( , , )
0 .
Solution: We have =−
=−
=−
u xy z
v yz x
w zx y
, , . Therefore,
= − −u x y zlog log log( ) (1)
= − −v y z xlog log log( ) (2)
= − −w z x ylog log log( ) (3)
Differentiating (1) partially w.r.t. x, we get
⋅ ∂∂
=u
ux x
1 1 or
∂∂
=ux
ux
Differentiating (1) partially w.r.t. y, we get
⋅ ∂∂
= −−u
uy y z
1 1 or ∂∂
= −−
uy
uy z
Differentiating (1) partially w.r.t. z, we get
⋅ ∂∂
= −−
−u
uz y z
1 1 ( 1) or ∂∂
=−
uz
uy z
Similarly, from (2) and (3), we have
∂∂
=−
∂∂
= ∂∂
= −−
∂∂
= −−
∂∂
=−
∂∂
=vx
vz x
vy
vy
vz
vz x
wx
wx y
wy
wx y
wz
wz
, , , , ,
\ ∂∂
=
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
u v wx y z
ux
uy
uz
vx
vy
vz
wx
wy
wz
( , , )( , , )
=
−− −
−−−
−− −
ux
uy z
uy z
vz x
vy
vz x
wx y
wx y
wz
Functions of Several Variables 99
Taking out common u, v, w from R1, R2, R3 respectively, we get
u v wx y z
( , , )( , , )∂∂
=
−− −
−−−
−− −
uvw
x y z y z
z x y z x
x y x y z
1 1 1
1 1 1
1 1 1
Multiplying R1, R2, R3 by y – z, z – x, x – y, respectively, we get
u v wx y z
( , , )( , , )∂∂
=− − −
− −
− −
− −
uvwy z z x x y
y zx
z xy
x yz
( )( )( )
1 1
1 1
1 1
Multiplying C1, C2, C3 by x, y, z, respectively, we get
u v wx y z
( , , )( , , )∂∂
=− − −
− −− −
− −
uvwxyz y z z x x y
y z y zx z x zx y x y( )( )( )
Operating C1 C1 + C2 + C3, we get
u v wx y z
( , , )( , , )∂∂ =
− − −
−− −
−=
y z z x x y
y zz x zy x y
1( ) ( ) ( )
000
02 2 2
2.5.3 Functional Relationship
Let u u u, ,1 2 3 be the functions of x x x, ,1 2 3, respectively. Then the functional relationship of the form
=f u u u( , , ) 01 2 3 exists if
=J
u u ux x x
, ,, ,
01 2 3
1 2 3 and conversely, if the functional relationship of the form
=f u u u( , , ) 01 2 3 exists, then
=J
u u ux x x
, ,, ,
01 2 3
1 2 3.
Example 2.28 Show that the functions = + +u x y z , = + + −v x y z xyz33 3 3 and
= + + − − −w x y z xy yz zx2 2 2 are functionally dependent and find the relation between
them.
Solution: The functions u, v, w are functionally dependent if ∂∂
=u v wx y z
( , , )( , , )
0 . Therefore,
100 Engineering Mathematics
∂∂
=
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
= − − −− − − − − −
u v wx y z
ux
uy
uz
vx
vy
vz
wx
wy
wz
x yz y xz z xyx y z y z x z x y
( , , )( , , )
1 1 13( ) 3( ) 3( )2 2 2
2 2 2
= − − + − − + −− − − −x yz y x z y x z x y z xx y z y x z x
31 0 0
( ) ( )2 3( ) 3( )
2 2 2 2 2
Expanding along R1, we get
u v wx y z
( , , )( , , )∂∂
=− + + − + +
− −= − − + + =
y x x y z z x x y zy x z x
y x z x x y z9( )( ) ( )( )
9( )( )( ) 1 11 1
0
Therefore, u, v, w are functionally dependent.
Now, = + + − = + + + + − − − =v x y z xyz x y z x y z xy yz zx uw3 ( )( )3 3 3 2 2 2 .Hence, =v uw is the desired relation.
Example 2.29 If = = =u yzxv zx
yw xy
z, , show that ∂
∂=u v w
x y z( , , )( , , )
4
Solution:
∂∂
=
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
u v wx y z
ux
uy
uz
vx
vy
vz
wx
wy
wz
( , , )( , , )
=
−
−
−
=
−−
−
yzx
zx
yx
zy
zxy
xy
yz
xz
xyz
x y z
yz xz xyzy zx xyyz xz xy
1 1 1
2
2
2
2 2 2
= − −
−
− −
+ +
= + + =yz
xx yzy z
xyz
zx
xyyz
xz
yxxyxy
0 2 2 42
2
2 2
2
Functions of Several Variables 101
Example 2.30 Prove u = x + y + z; v = xy + yz + zx; w = x2 + y2 + z2 are functionally independent. Find the relationship between them.
Solution:
∂∂
=
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
u v wx y z
ux
uy
uz
vx
vy
vz
wx
wy
wz
( , , )( , , )
= + + +y z z x x yx y z
1 1 1
2 2 2
= + − + − + − + + + − + z z x y x y z y z x x y y y z x z x1 2 ( ) 2 ( ) 1 2 ( ) 2 ( ) 1 2 ( ) 2 ( )
= + − − − − + + + + − − =z xz xy y yz z x xy y yz xz x2 2 2 2 2 2 2 2 2 2 2 2 02 2 2 2 2 2
u,v and w are functionally dependent.The relation between them is given is (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) u2 = w + 2v
2.6 TAYLOR’S SERIES FOR FUNCTIONS OF TWO VARIABLES
Theorem 2.4 If f (x, y) possesses finite and continuous partial derivatives of all orders, then
+ + = + ∂∂
+ ∂∂
+
∂∂
+ ∂∂ ∂
+ ∂∂
+⋅ ⋅ ⋅f x h y k f x y h f
xk fy
h fx
hk fx y
k fy
( , ) ( , ) 12!
222
2
22
2
2
Proof: We know that Taylor’s theorem for a function f (x) of single variable x is
+ = + ′ + ′′ + ⋅ ⋅ ⋅f x h f x hf x h f x( ) ( ) ( )2!
( )2
Now, expanding + +f x h y k( , ) as a single variable x by keeping y as constant, we have
+ + = + + ∂∂
+ + ∂∂
+ +⋅⋅ ⋅f x h y k f x y k hxf x y k h
xf x y k( , ) ( , ) ( , )
2!( , )
2 2
2 (2.24)
and expanding + +f x h y k( , ) as a single variable y by keeping x as constant, we get
+ = + ∂∂
+ ∂∂
+⋅⋅ ⋅f x y k f x y kyf x y k
yf x y( , ) ( , ) ( , )
2!( , )
2 2
2 (2.25)
102 Engineering Mathematics
Using (2.25), (2.24) can be written as+ +f x h y k( , )
+ + = + ∂∂
+ ∂∂
+⋅⋅ ⋅
+ ∂
∂+ ∂
∂+ ∂
∂+⋅⋅ ⋅
f x h y k f x y kyf x y k
yf x y h
xf x y k
yf x y k
yf x y( , ) ( , ) ( , )
2!( , ) ( , ) ( , )
2!( , )
2 2
2
2 2
2
+ ∂∂
+ ∂∂
+ ∂∂
+⋅⋅ ⋅
hx
f x y kyf x y k
yf x y
2!( , ) ( , )
2!( , )
2 2
2
2 2
2+ ...
Hence, + + = + ∂∂
+ ∂∂
+
∂∂
+ ∂∂ ∂
+ ∂∂
+⋅ ⋅ ⋅f x h y k f x y h f
xk fy
h fx
hk fx y
k fy
( , ) ( , ) 12!
222
2
22
2
2 (2.26)
Symbolically, this result can be written as
+ + = + ∂∂
+ ∂∂
+ ∂
∂+ ∂
∂
+⋅ ⋅ ⋅+ ∂
∂+ ∂
∂
+⋅ ⋅ ⋅f x h y k f x y h
xkyf h
xky
fnhxky
f( , ) ( , ) 12!
1!
n2
Corollary: (i) Putting x = a and y = b, (2.26) becomes
+ + = + + + + + +⋅⋅ ⋅f a h b k f a b hf a b kf a b h f a b hkf a b k f a b( , ) ( , ) [ ( , ) ( , )] 12!
[ ( , ) 2 ( , ) ( , )]x y xx xy yy2 2
Putting a + h = x and b + k = y so that h = x – a, k = y – b, we get
= + − + −f x y f a b x a f a b y b f a b( , ) ( , ) [( ) ( , ) ( ) ( , )]x y
+ − + − − + − +⋅⋅ ⋅x a f a b x a y b f a b y a f a b12!
[( ) ( , ) 2( )( ) ( , ) ( ) ( , )]xx xy yy2 2 (2.27)
which is a Taylor’s expansion of f (x, y) in powers of (x – a) and (y – b). This is used for the expansion of f (x, y) in the neighbourhood of (a, b). (ii) Putting a = 0, b = 0 in (2.27), we get
= + +f x y f xf yf( , ) (0,0) [ (0,0) (0,0)]x y + + + +⋅⋅ ⋅x f xyf y f12!
[ (0,0) 2 (0,0) (0,0)]xx xy yy2 2 (2.28)
This is a Maclaurin’s expansion of f (x, y). This is used for the expansion of f (x, y) in the neighbourhood of origin (0, 0).
Example 2.31 Expand e ysinx in powers of x and y as far as terms of the third degree.
Solution: Let =f x y e y( , ) sinx or =f (0,0) 0
=f x y e y( , ) sinxx or =f (0,0) 0x ; =f x y e y( , ) cosy
x or =f (0,0) 1y
=f x y e y( , ) sinxxx or =f (0,0) 0xx ; =f x y e y( , ) cosxy
x or =f (0,0) 1xy
= −f x y e y( , ) sinyyx or =f (0,0) 0yy ; =f x y e y( , ) sinxxx
x or =f (0,0) 0xxx
=f x y e y( , ) cosxxyx or =f (0,0) 1xxy ; = −f x y e y( , ) sinxyy
x or =f (0,0) 0xyy
Functions of Several Variables 103
= −f x y e y( , ) cosyyyx or = −f (0,0) 1yyy
Maclaurin’s expansion of f x y( , ) is given by
f x y( , ) = + + + + +f x f y f x f xy f y f(0,0) [ (0,0) (0,0)] 12!
[ (0,0) 2 (0,0) (0,0)]x y xx xy yy2 2
+ + + + +⋅⋅ ⋅x f x y f xy f y f13!
[ (0,0) 3 (0,0) 3 (0,0) (0,0)]xxx xxy xyy yyy3 2 2 3
= + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + − +⋅⋅ ⋅x y x xy y x x y xy y0 ( 0 1) 12!
( 0 2 1 0) 13!
[ 0 3 1 3 0 ( 1)]2 2 3 2 2 3
= + + − +⋅⋅ ⋅y xy x y y12
16
2 3
Example 2.32 Expand + −x y y3 22 in powers of (x – 1) and ( y + 2) using Taylor’s theorem.
Solution: Let = + − ⇒ − = −f x y x y y f( , ) 3 2 (1, 2) 102 . Therefore,
= − = − = + − =f x y xy f f x y x f( , ) 2 or (1, 2) 4; ( , ) 3 or (1, 2) 4;x x y y2
=f x y y( , ) 2 orxx − = − = − = = − =f f x y x f f x y f(1, 2) 4; ( , ) 2 or (1, 2) 2; ( , ) 0 or (1, 2) 0xx xy xy yy yy
− =f (1, 2) 0;xxx − = − = − =f f f(1, 2) 2; (1, 2) 0; (1, 2) 0xxy xyy yyy .
All partial derivatives of higher order vanish.Taylor’s expansion of f x y( , ) in powers of −x a( ) and ( y – b) is given by
= + − + − + −f x y f a b x a f a b y b f a b x a f a b( , ) ( , ) [( ) ( , ) ( ) ( , )] 12!
[( ) ( , )x y xx2
+ − − + − + −x a y b f a b y b f a b x a f a b2( )( ) ( , ) ( ) ( , )] 13!
[( ) ( , )xy yy xxx2 3
+ − − + − −x a y b f a b x a y b f a b3( ) ( ) ( , ) 3( )( ) ( , )xxy xyy2 2 + − +⋅⋅ ⋅y b f a b( ) ( , )]yyy
3 (1)
+ − = − + − − + + + − − + − + + +x y y x y x x y y3 2 10 [( 1)( 4) ( 2)4] 12
[( 1) ( 4) 2( 1)( 2)(2) ( 2) (0)]2 2 2
+ − + − + + − + + +x x y x y y16
[( 1) (0) 3( 1) ( 2)(2) 3( 1)( 2) (0) ( 2) (0)]3 2 2 3
= − − − + + − − + − + + − +x y x x y x y10 4( 1) 4( 2) 2( 1) 2( 1)( 2) ( 1) ( 2)2 2
Example 2.33 Expand +e ylog(1 )x in powers of x and y upto the terms of third degree.
Solution: Here, = +f x y e y( , ) log(1 )x or =f (0,0) 0 ; = +f x y e y( , ) log(1 )xx or =f (0,0) 0x
=+
f x y ey
( , ) 11y
x or =f (0,0) 1y ; = +f x y e y( , ) log(1 )xxx or =f (0,0) 0xx ;
104 Engineering Mathematics
=+
f x y ey
( , ) 11xy
x or =f (0,0) 1xy ; = − + −f x y e y( , ) (1 )yyx 2 or = −f (0,0) 1yy ;
= +f x y e y( , ) log(1 )xxxx or =f (0,0) 0xxx ; =
+f x y e
y( , ) 1
1xxyx or =f (0,0) 1xxy
= − + −f x y e y( , ) (1 )xyyx 2 or = −f (0,0) 1xyy ; = + −f x y e y( , ) 2 (1 )yyy
x 3 or =f (0,0) 2yyy
Maclaurin’s expansion of f (x, y) is given by
= + + + + +f x y f xf yf x f xy f y f( , ) (0,0) { (0,0) (0,0)} 12!
{ (0,0) 2 (0,0) (0,0)}x y xx xy yy2 2
+ + + + +⋅⋅ ⋅x f x y f xy f y f13!
{ (0,0) 3 (0,0) 3 (0,0) (0,0)}xxx xxy xyy yyy3 2 2 3
\ + = + + + + + −e y x y x xy ylog(1 ) 0 (0) (1) 12!
{ (0) 2 (1) ( 1)}x 2 2
+ + + − + +⋅⋅ ⋅x x y xy y13!
{ (0) 3 (1) 3 ( 1) (2)}3 2 2 3
= + − + − + +⋅⋅ ⋅y xy y x y xy y12
12
( ) 13
2 2 2 3
Example 2.34 Find the expansion of cos xcos y in powers of x, y up to fourth-order terms.
Solution: Let =f x y x y( , ) cos cos or = = = − =f f x y x y f(0,0) cos0cos0 1; ( , ) sin cos or (0,0) 0x x
= − =f x y x y f( , ) cos sin or (0,0) 0y y ; = − = −f x y fcos cos or (0,0) 1xx xx
= =f x y x y f( , ) sin sin or (0,0) 0xy xy ; = − = −f x y x y f( , ) cos cos or (0,0) 1yy yy
= =f x y x y f( , ) sin cos or (0,0) 0xxx xxx ; = =f x y x y f( , ) cos sin or (0,0) 0xxy xxy
= =f x y x y f( , ) sin cos or (0,0) 0xyy xyy ; = =f x y x y f( , ) cos sin or (0,0) 0yyy yyy
= =f x y x y f( , ) cos cos or (0,0) 1xxxx xxxx ; = − =f x y x y f( , ) sin sin or (0,0) 0xxxy xxxy
= =f x y x y f( , ) cos cos or (0,0) 1xxyy xxyy ; = − =f x y x y f( , ) sin sin or (0,0) 0xyyy xyyy
= =f x y x y f( , ) cos cos or (0,0) 1yyyy yyyy
Maclaurin’s expansion of f x y( , ) is given by
f x y( , ) = + + + + +f xf y f x f xy f y f(0,0) [ (0,0) (0,0)] 12!
[ (0,0) 2 (0,0) (0,0)]x y xx xy yy2 2
+ + + +x f x y f xy f y f13!
[ (0,0) 3 (0,0) 3 (0,0) (0,0)]xxx xxy xyy yyy3 2 2 3
+ + + + +x f x y f x y f xy f y f14!
[ (0,0) 4 (0,0) 6 (0,0) 4 (0,0) (0,0)]xxxx xxxy xxyy xyyy yyyy4 3 2 2 3 4 + …
Functions of Several Variables 105
\ = + ⋅ + ⋅ + − + ⋅ + − + ⋅ + ⋅ + ⋅ +x y x y x xy y x x y xy ycos cos 1 ( 0 0) 12!
[ ( 1) 2 0 ( 1)] 13!
( 0 3 0 3 0 )2 2 3 2 2 3
+ ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ ⋅ ⋅x x y x y xy y14!
[ 1 4 0 6 1 4 0 1]4 3 2 2 3 4
= − − + + + +⋅⋅ ⋅x y x x y y12! 2! 4!
64!
14!
2 2 4 2 24
= − + + + + +⋅⋅ ⋅x y x x y y1 12
( ) 124
( 6 )2 2 4 2 2 4
Example 2.35 Expand x3 + y3 + xy2 in the power of x 1 & y 2 upto 3 degree terms .
Solutions:
Let f (x, y) = x3 + y3 + xy2
Let f (x, y) = x3 + y3 + xy2 f (1, 2) = 1 + 8 + 4 = 13
fx (x, y) = 3x2 + y2 fx (1, 2) = 3 + 4 = 7
fy (x, y) = 3y2 + 2xy fy (1, 2) = 12 + 4 = 16
fxx (x, y) = 6x fxx (1, 2) = 6
fxy (x, y) = 2y fxy (1, 2) = 4
fyy (x, y) = 6y + 2x fyy (1, 2) = 6 + 4 = 10
fxxx (x, y) = 6 fxxx (x, y) = 6
fxxy (x, y) = 0 fxxy (x, y) = 0
fxyy (x, y) = 2 fxyy (x, y) = 2
fyyy (x, y) = 6 fyyy (x, y) = 6
( ) ( )
= + − + − + − + − − + −
+
− + + − − + −
f x y x y x x y y
x x y y
( , ) 13 ( 1)7 ( 2)16 12!
6( 1) 8( 1)( 2) 14( 2)
13!
6( 1) 0 6 1 2 6( 2)
2 2
3 3
= + − + − + − + − − + −
+ − + − − + −
f x y x y x x y y x x y y( , ) 13 7 7 16 32 3( 1) 4( 1)( 2) 7( 1) ( 1) ( 1)( 2) ( 2)2 2 3 3
Example 2.36 Expand the function sinxy in power of x 1 and π−y2
upto second degree terms.
Solutions: Let f (x, y) = sin xy π
=f 1,
20
x
π
=f 1,
20
x 1
=f x y y xy( , ) cos( )x π
=f 1,
20
x
106 Engineering Mathematics
=f x y x xy( , ) cos( )y
π
=f 1,
20
y
= −f x y y xy( , ) sin( )xx
2
π π
= −f 1,
2 4xx
2
= −f x y x xy( , ) sin( )yy
2
π
= −f 1,
21
yy
= − +f x y xy xy xy( , ) sin( ) cos( )xy
π π
=f 1,
2 2xx
( )π π π π= + − + −
+ − −
−
+ − −
xy x y x y ysin 1 ( 1)02
0 12!
( 1)2 2 2
( 1)2
π π π π= + −− − − −
− −
x x y y1 12 4
( 1) ( 1)2 2
22
2
Example 2.37 Expand e ycosx about π
0,
2upto the third term using Taylor’s series.
Solutions: Let f (x, y) = e ycosx
π
π
π
π
π
π
π
π
=
=
= −
= −
=
=
= −
=
=
=
= −
= −
= −
= −
=
=
= + + − + − +
f f
f f
f f
f f
f f
f f
f f
f f
x y e y
x y e y
x y e y
x y e y
x y e y
x y e y
x y e y
x y e y
e y xx y x x y
( , ) cos 0,2
0
( , ) sin 0,2
1
( , ) cos 0,2
0
( , ) cos 0,2
0
( , ) cos 0,2
0
( , ) sin 0,2
1
( , ) sin 0,2
0
( , ) sin 0,2
1
cos 11!
12!
( ) 13!
( 3 ) ...
x
x
x
y
x
y
xx
x
xx
yy
x
yy
xxx
x
xxx
xxy
x
xxy
xyy
x
xyy
yyy
x
yyy
x 2 2 3 2
π
π
π
π
π
π
π
π
=
=
= −
= −
=
=
= −
=
=
=
= −
= −
= −
= −
=
=
= + + − + − +
f f
f f
f f
f f
f f
f f
f f
f f
x y e y
x y e y
x y e y
x y e y
x y e y
x y e y
x y e y
x y e y
e y xx y x x y
( , ) cos 0,2
0
( , ) sin 0,2
1
( , ) cos 0,2
0
( , ) cos 0,2
0
( , ) cos 0,2
0
( , ) sin 0,2
1
( , ) sin 0,2
0
( , ) sin 0,2
1
cos 11!
12!
( ) 13!
( 3 ) ...
x
x
x
y
x
y
xx
x
xx
yy
x
yy
xxx
x
xxx
xxy
x
xxy
xyy
x
xyy
yyy
x
yyy
x 2 2 3 2