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CHAPTER 11
PARTIAL DIFFERENTIATION
TOPICS:
1. DEFINITION, FIRST AND SECOND ORDER PARTIAL DERIVATIVES 2.HOMOGENEOUS FUNCTIONS AND EULARS THEOREM.
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PARTIAL DIFFERENTIATION
Let u
(i)
f ( x , y ) be a function of two independent varaibles x and y.
f ( x
If Lt h
h, y )
f ( x, y )
exists then the limit is called the partial derivatie of u with
h
0
u
respect to x. It is denoted by
(ii) If Lt k
f ( x , y
k)
f ( x, y )
u
respect to y. It is denoted by
∴
∂u ∂ x
= Lt
f ( x + h , y ) − f (x , y )
h→0
f
or
.
x
exists then the limit is called the partial derivative of u with
k
0
x
or u x or
h
or u y or
and
∂u ∂ y
f
or f y .
= Lt
f ( x, y + k ) − f ( x, y ) k
k →0
Note : (i)The partial derivative of u w.r.t. x is the ordinary derivative of u w.r.t. x treating the other variable y (and its functions) as constant
(ii) The partial derivative of u w.r.t. y is the ordinary derivative of u w.r.t. y treating the other variable x (and its functions) as constant.
DIFFERENTIATION OF COMPOSITE FUNCTIONS
1.
If V = g(U) and U = f(x, y) then (i)
2. If Z = f(x, y) and x = g(t); y = h(t) then
V
dV
x
dU
dZ dt
.
U , x
(ii)
Z dx . x dt
Z dy y dt
V
dV
y
dU
.
U y
is called the total differential
coefficient of Z w.r.t. t. 3. If f(x, y) = c where c is constant, then
dy
f
f
dx
x
y
.
PARTIAL DERIVATIVES OF SECOND ORDER Definition : If U = f(x, y) then
U x
,
U y
are called the partial derivatives of first order and they
are functions of x, y. The partial derivatives of
U x
and
U y
, if they exist, are called the second
order partial derivatives . They are denoted by www.sakshieducation.com
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U x x
U
U xx
x2
U and y y
f xx
U , y x
2
U
y x
U yx
f yx ,
U x y
2
U
x y
U
f xy
y
2
U
U yy
y2
f yy .
HOMOGENEOUS FUNCTION:
A function u
f ( x , y ) is said to be a homogeneous function of degree n in the variables in x k n f ( x , y ) for all k or f ( x , y )
and y if f ( kx , ky )
y or f ( x , y ) x
x n f
x .
y n f
EULER’S THEOREM
If u
f ( x , y ) is a homogeneous function of degree n in the variables x,y then x
∂U ∂ x
+y
∂U ∂y
= nU
Proof:
Since u f ( x , y ) is a homogeneous function of degree n, we have U = xn g ( y / x ) where g ( y / x ) is function of y/x. ∴
∂U ∂ x
and ∴ x
−y n −1 + nx .g ( y / x ) 2 x
n
= x .g ′( y / x )
∂U ∂ x
∂U ∂ y +y
-- (1)
1 x
n
= x .g ′( y / x ) ∂U ∂y
y n n −1 + nx g ( y / x) + yx . g ′( y / x) 2 x
n
= x x g ′( y / x) −
= n.x n .g ( y / x ) == nU.
Note 1: If U x
∂U ∂ x
+y
∂U
= f ( x , y, z )
+z
∂y
∂U ∂z
is a homogeneous function of degree n in x,y,z then
= nU
Therorem If U = f ( x, y ) is a homogeneous function of degree n in x,y then x
2
2
∂ U ∂ x
2
2
+ 2 xy
∂ U ∂ x∂y
+y
2
2
∂ U ∂y
2
= n (n − 1)U .
Proof:
Since U = f ( x, y ) is a homogeneous function of degree n, by Euler’s theorem, we have
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x
∂U ∂ x
+y
∂U
= nU ------(1)
∂y
Differentiating (1) partially w.r.t. x we get 2
2
∂U x 2 + ∂ x ∂ x
∂ U .1 + y ∂x∂y
∂ U
2
∂U
2
∂U ∂U ∂U =n +y = (n − 1) -- (2) ⇒ x 2 ∂x ∂ x∂y ∂x ∂ x
Differentiating (1) w.r.t. y partially we get, 2
x
2
∂ U
+y
∂ y∂x
∂ U ∂ y
2
+
(2) . x + (3) . y
∂U ∂y ⇒
.1 = n
x
2
∂U ∂y
⇒
2
∂ U ∂ x
2
2
x
∂U ∂ y∂x
2
+y
2
+ 2 xy
∂ U ∂ x∂y
+y
2
∂U ∂ y
2
= (n − 1)
∂U ∂y
-------(3)
2
∂ U ∂y
2
= n (n − 1)U .
PARTIAL DIFFERENTIATION EXERCISE – 11(a) I. 1.
z
Find
x
i)
z
3xe
z
,
for
y
y2
4y
differentiate partially w.r.t x, Sol:
∂z ∂x
= 3e
∂
y2
∂x
(x ) +
∂ ∂x
( 4 y ) = 3e y
2
differentiate partially w.r.t y, ∂z ∂y
=
∂ ∂
(3 x e ) y y2
z
log y
Sol: z
log y
ii)
+
∂ ∂y
(4 y ) =
2
6 x ye y + 4
x
2 y x
y 2
differentiate partially w.r.t x, ∂z ∂x
=
1 y+
x y + 2 x ∂x y
y2
∂
=
1 1 = y3 + x y 2 y3 + x y2
differentiate partially w.r.t y,
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y2 −2 1 x = + 3 y3 + x y
x = y + ∂y y 2 x ∂y y + 3 y ∂z
=
iii)
∂
1
y2 3 y +x
3 y − 2x ) ( = y ( y3 + x )
y3 − 2 x 3 y
y2 Tan x 1
z
Sol: differentiate partially w.r.t x,
y2 = 2 ∂x x ∂x 2 y 1+ x ∂z
∂
1
=
2 − y2 −y = x 2 + y 4 x 2 x 2 + y4
x2
differentiate partially w.r.t y,
y2 . = ∂y y 2 ∂y x 1+ x ∂z
iv)
z
∂
1
cosx
ans: ∂ z
s in y
∂x
− s in x
=
&
s in y
∂z ∂y
=
− cos x 2
sin y
( cosy )
= − c o s x . co t y .c o s e c y
v) vi)
xey
z
ans. ∂z
yex .
∂x
=e
y
+ ye
∂z
x,
∂y
= xe
y
+e
x
1
z
1 x
y
2
Sol: differentiate partially w.r.t x, ∂z ∂x
=
−1
2
(1 + x + y 2 )
−3 / 2
∂
1+ x + y2 ) ( ∂x
=
−1
2
(1 + x + y 2 )
−3 / 2
differentiate partially w.r.t y, ∂z ∂y
=
= −
−1
( 2
1
1 + x + y2 ) ( 2
1 + x + y2
)
−3 / 2
−3 / 2
∂ ∂y
( 2y ) =
(1 + x + y 2 ) −y 3/ 2
(1 + x + y 2 )
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sin x2
(
)
[ans. = 2x cos x2 − y , − cos ( x 2 − y )
vii)
x
2.
For the following functions f, show that f xx
i)
f x
ii)
e x sin y
x
y
2
y
f yy
0.
2
x
Sol: f = e sin y
Differentiate f partially w.r.t x, f x = e x sin y , differentiate f x partially w.r.t x, fxx = ex sin y f = e x sin y differentiate f partially w.r.t y, f y = e x cos y, differentiate f x partially w.r.t y, x
∴ f yy = − e sin y ∴ f xx + f yy = 0
iii)
f= s i n
3.
If v
4.
If z
Sol: z
x . co sh y
r 2h , show that rvr
sin x
sin x
y
y
log x log x
2hvh
4v
y show that zxx
zyy .
y
Differentiate partially w.r.t x, z x = cos ( x − y ) +
1 x + y
Again differentiate partially w.r.t x, 1 z xx = − sin ( x − y ) − 2 ( x + y)
---(1)
Differentiate z partially w.r.t y, z y = cos ( x − y ) +
1 x + y
Aganin differentiate partially w.r.t y,
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z yy = − sin ( x − y ) − ∴
5.
1
( x + y)
---(2)
2
From (1) and (2), we get zxx = zyy
If u 3 1 a3
8 x ay b
3
then show that u 3x
u 3y
8
3
3
Sol: u =
8 ( x + ay + b ) 1 + a3 2
u=
3 1+ a
( x + ay + b )
3
differentiate partially w.r.t x
differentiate partially w.r.t y
6.
If au
7.
If z
Sol: z
2
b
Ae
Ae
p2t
u y =
3 1+ a
3
2a
3
3 1 + a3
y , then show that uxuy
a x p2t
⇒
2
u x =
⇒
cospx , then prove that z xx
8
3
∴ ux + uy =
1 + a3
1
zt
cospx
differentiate partially w.r.t x, z x = A.e − p = − Ap.e
2
2
t
−p t
( − p sin px ) sin Px
Again differentiate partially w.r.t x 2
z xx = −Ap 2 .e − p t .cos px
---(1)
differentiate z partially w.r.t t, z t = A cos px.e − p
2
t
(
−p
2
)
2 − p2 t
= − Ap e
cos px
---(2) From (1) and (2) we get z xx = z t
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+
83 1+ a3
=
(
8 1 + a3 1+ a3
) =8
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II. 1.
Find all the first and second order partial derivatives of the following functions f. i) sin (xy) 1
ii) tan tan
x
tan
1
y
iii) e x cos y y
iv) ex i)
z = sin (xy) differentiate partially w.r.t x, z x = cos ( xy ) .y = xy cos ( x.y )
Again differentiate partially w.r.t x, z xx = − y sin ( xy ) .y = − y 2 sin ( xy )
Differentiate z partially w.r.t y, z y = ( cos xy ) .x = x cos xy
Again differentiate partially w.r.t x, z xy = ( z y ) = x ( − sin xy.y ) + cos xy x
= − xy sin ( xy ) + cos ( xy )
Differentiate z y partially w.r.t y, 2
z yy = − x sin ( xy ) .x = − x sin ( xy )
( z x )y
= ( y cos xy )
(
y
)
= y sin ( xy ) x + cos xy = cos xy − x.y.sin ( xy )
ii)
z= tan tan
iii)
e x cos y
1
x
tan
1
y
Sol: z = ex cos y
differentiate partially w.r.t x, x
z x = e cos y, again diff. partially w.r.t x, x z xx = e cos y
.
differentiate z partially w.r.t y,
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z y = −e sin y,
again diff. partially w.r.t y, x z yy = −e cos y
Differentiate z y partially w.r.t x, z yx = ( z y ) = −e x .sin y x
z x = ex cos y z y = z xy = z yx = − e x .sin y
ex
y
iv)
z
2.
For the following functions f, show that f xx
i)
0.
f yy
y x2
y2 y
Sol: f = x
2
y
differentiate partially w.r.t x, f x
2
− y ( 2x )
=
(x2 + y2 )
2
Again differentiate partially w.r.t x,
(
2
−2y x + y
f xx =
2
)
(
(
(
x 2 + y2
3
)
2
+ 2xy.2 x + y
x 2 + y2
2 − x 2 − y 2 + 4x 2
=
2
2
).2x
4
=
(
)
2y x 2 + y 2 − x 2 − y 2 + 4x 2
( x 2 + y2 )
4
) = 2y (3x 2 − y2 ) 3 2 2 + x y ( )
Differentiate f partially w.r.t y, f y =
(
)
1 x 2 + y 2 − 2y 2
(
2
x +y
2
)
x 2 − y2
=
2
(
x2 + y2
)
2
differentiate partially w.r.t y,
(
2
−2y x + y
f yy =
=
2
)
2
(
2
( x2 + y2 )
(
2y y 2 − 3x 2
(x
2
+y
2
)
3
)
2
−2 x + y
)( x2 − y2 ) ( 2y)
4
=
(
) 4 2 2 + x y ( )
2y x 2 + y 2 − x 2 − y 2 − 2x 2 + 2y 2
∴ f xx + f yy = 0
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y x
ii)
Tan
Sol:
f = ta n − 1
y x
differentiate partially w.r.t x, 2
1 −x y −y fx = . = 2 2 2 2 y x x +y x 1+ x 1
=
−y
x2 + y2
differentiate partially w.r.t x, f xx =
+ y ( 2x )
( x 2 + y2 )
2
differentiate f partially w.r.t y, 1 1 x fy = . = 2 2 2 y x x +y 1+ x differentiate partially w.r.t y,
iii)
f= log x 2
iv)
e
x
y
x sin y
Sol: f = e
x
f yy =
−2xy
(x
2
+y
2
)
∴ f xx + f yy = 0
2
2
y cos y
x sin y
y cos y
differentiate partially w.r.t x, f ( x ) = e− x ( x sin y − y cos y ) f x = e − x ( x sin y − y cos y ) + e − x ( sin y )
differentiate partially w.r.t x, f xx = e − x ( x sin y − y cos y ) − e − x .sin y − e − x .sin y = e− x ( x sin y − y cos y − 2sin y )
differentiate f partially w.r.t y, x f y = e − ( x cos y − cos y + y sin y )
differentiate partially w.r.t y, −x f yy = e ( − x sin y + sin y + sin y + y cos y ) = e
−x
( −x sin y + y cos y + 2sin )
∴ f xx + f yy = 0 x
v)
e
vi)
e 2xy cos y 2
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vii)
e2x A sin 2y B cos 2y
Sol: f= e2x A sin 2y B cos 2y Diff. f partially w.r.t x, f x = 2e 2x ( A sin 2y + B cos 2y ) f xx = 4e 2x ( A sin 2y + B cos 2y )
Again diff. partially w.r.t x, Diff, f partially w.r.t. y, f y
=e
2x
Again diff. partially w.r.t. y, = −4e ⇒
f yy = e
2x
( − 4A sin 2y − 4B cos 2y )
( A sin 2y + B cos 2y ) =−f xx
f xx + f yy = 0
viii) f= e
3.
2x
( 2A cos 2y − 2B sin 2y )
2xy
sin x
2
y
2
x2
If r, , x and y are connected by the equations r rx , ry ,
x
y.
and
Also verify that ry
y
rx
y2
0.
x
2 2 Sol: r = x + y
differentiate partially w.r.t x, 1 x rx = .2x = 2 x 2 + y2 x 2 + y2 differentiate r partially w.r.t y, 1 y ry = .2y = 2 x 2 + y2 x 2 + y2 θ = ta n
−1
y x
differentiate partially w.r.t x,
θx =
y − 2 x2 y 1+ x 1
=−
y x 2 + y2 ) 2( x
=−
y 2
x +y
2
x2
differentiate partially w.r.t y, θy =
1
y 1+ x
2
.
1 x
=
x x 2 + y2
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1/ 2
;
Tan
1
y . find x
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y
ry .θx + rx .θ y =
4.
If z
Sol: z
tan y tan y
2
x +y
ax
2
y
ax
y
.
x x 2 + y2
ax
ax
+
1/ 2
, find z xx
y xy − xy − 2 = 2 x 2 + y2 x + y x 2 + y2 x
a 2 z yy .
1/ 2
Differentiate partially w.r.t x, z x = a sec 2 ( y + ax ) +
1 2
( − a ) ( y − ax )
−1 / 2
Differentiate partially w.r.t x, z xx = 2a 2 sec 2 ( y + ax ). tan ( y + ax ) −
a2 4
( y − ax )
−3/ 2
Differentiate z partially w.r.t y, z y = sec 2 ( y + ax ) +
1 2
( y − ax )
−1 / 2
Differentiate partially w.r.t y, z yy = 2 sec 2 ( y + ax ) tan ( y + ax ) − 2
2
∴ z xx − a z yy = 2a sec
2
1 4
( y − ax )
−3 / 2
( y + ax ) tan ( y + ax ) 2
−2a sec
2
−
a2 4
( y − ax )
( y + ax ) tan ( y + ax ) +
−3/ 2
a2 4
( y − ax )
2
∴ z xx − a z yy = 0
5.
If 2z ay2
6.
If z a ex
7.
If u
2b
ay
2
16ax , show that zy
b , then show that z x z
x
2
y
2
zx 2
1
2
2
xyzx
z
2
, then show that
∑
u
x
2
zy
0.
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−3/ 2
=0
=0
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Exercise – 11(b) I. 1.
Which of the following are homogeneous functions?
i)
f x, y
x1/ 3 .y 3 / 4 , tan
Sol: f x, y
x1/ 3 .y 3 / 4 , tan
= k
1/3
1
y x
1
y f ( k x , k y ) = ( k x )1 / 3 ( k y )3 / 4 . ta n − 1 k y x kx y x
.x 1 / 3 .k 3 / 4 .y 3 / 4 ta n − 1
1 3
1
+
3
= k 3 4 .x 3 .y 4
y x
tan −1
13
(x, y )
= k 12 f
f ( x, y ) is a homogeneous function.
ii)
y x
3x
f x, y
log
y
Ans ; f ( x, y ) is a homogeneous function. iii)
f x, y
log y
2 log x
Sol: f ( kx1 ky ) = log ky + 2 log kx = log k + log y + 2 ( log k + log x ) = 3 log k + ( log y + 2 log x ) = 3log k + f ( x, y )
f ( x, y ) is not a homogeneous function.
iv)
x y y.g x y
u x, y
xf
Ans : homogeneous function. v)
x3
f x, y
y3
3/2
Ans. f is a homogeneous function. vi)
1 / 3 1/ 3
f x, y
x
y
1/ 3
Sol:
f ( kx, ky ) = ( kx ) =k
2/3 1/ 3 1/ 3
=k
.x
2/3
.y
x
2/ 3
1/ 3
( ky )
.y
1/ 3 2/ 3
+ ( kx )
1/ 3
. ( ky )
. + k.x2/ 3 .y1.3
( x1/ 3.y1/ 3 + k1/ 3.x 2 / 3.y1/ 3 )
n
= k f ( x, y )
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f ( x, y ) is not a homogeneous function.
II. 1.
Verify Euler’s theorem for the following.
i)
f(x,y)
Sol. f(x,y)
x
y
x
y
x
y
x
y
Differentiate partially w.r.t 1 1 1 (x) −1/ 2 x + y − (x) −1/ 2 ( x − y (x) −1/ 2 x + y − x + y 2 ∂f 2 2 x, = = 2 2 ∂x x + y x + y
∂f ∂y
x
=
∂y
y
+
(xy)1/ 2
=
2
y
∂y
x
{
x− y
}
− =
1 2
(y)−1/ 2 2 x
1/ 2
x ∂ f ∂ x
+y
y
+
∂f ∂y
2
=o
Degree of the given function is 0. ∂ f ∂f +y = nf = 0. f = 0 x ∂ x ∂y Hence f(x, y) =
x− y x+ y
x + y
− (xy)
=
2
x + y Differentiate partially w.r.t y, 1 −1 (y) −1/ 2 x + y − (y) −1/ 2 2 ∂f 2 = 2 ∂y x + y ∂f
(x) −1/ 2 ( y)1/ 2
x
∂f
holds Euler’s theorem.
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2
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ii)
f ( x , y ) = tan −
y
1
x
2
iii)
Sol.
f ( x, y) = f ( x, y ) =
x y x
3
y
x 2 y x3 + y 3
3
k3
→ f ( kx, ky ) =
x2 y
k 3 x3 + y 3
x2 y
0
= k
x3 + y3
0
= k f
f is a homogeneous function of degree 0. ∴x
∂f ∂x
+
∂ f
ay
= 0.f = 0
differentiate f partially w.r.t x, ∂f ∂x
2xy(x 3 + y 3) − 3x 2 .x2 y
=
3
3 2
(x + y )
x
∂f ∂x
=
2x 2y(x 3 + y3 ) − 3x5 .y (x 3 + y3 )2
differentiate f partially w.r.t y, ∂f ∂y
y
=
∂f ∂y
∴x
x 2 (x3 + y3 ) − 3y2.x2 y (x3 + y3 )2 yx 5 + y4 x 2 − 3y4 x 2
=
∂f ∂x
(x 3 + y3 )2 +
∂ f ∂y
= 0.f
=0
Hence eulers theorem verified. 1
y
x/ y xe x
iv)
f (x, y) = x tan
2)
y y If u = x y then show that x 2u xx x x
x3 y 3 3) If u tan then show that xu x x3 y 3 x 3 + y3 Sol. u = tan −1 x 3 + y3 1
tan u =
2xy u xy
yu y
y 2u yy
0 (May.’06)
x 3 − y3 x 3 + y3
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0
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let tan u = z ( x , y ) = =k
0
x 3 − y 3 3
x + y
3
x 3 − y 3
( kx , ky ) =
→ f
3 3 x + y
k 3 x3 − y3 k
3
3 3 x +y
0
=k z
tanu is a homogeneous function degree Zero in x and y ∂ z ∂z +y = nz x ∂ x ∂y
x
∂ ∂x
∂
(ta n u ) + y
x.sec 2 u.
∂u ∂x
sec 2 u x.
(ta n u ) = 0 . ta n u .
∂y
+ y. s e c
∂u ∂x
+
2
y.
∂u
= 0
∂y
=0 ∂y
∂u
⇒
x.
∂u ∂x
+ y.
∂u ∂y
= 0
i.e., x.u u + y.u y = 0
PROBLEMS FOR PRACTICE 1.
If z = eex sin by, where a and b are real Constants, find ZZ,Zy , Zxx , Zxy , ZyyandZyz .
2.
Find all the first and second order partial Derivatives for f(x, y) = ex
3.
Find all the first and second order partial derivatives for f(x, y) = sin (ax + by) where a and b are real constants.
2y
2
x y
4.
If u (x, y) =
5.
If z = log (tan x + tan y), show that (Sin2x)z z
6.
If u
e xy , show that u u xx
7.
If z
f x
8. 9.
If u
3xy
If f x, y
2
x
3
y y3 2
2
y
3
, show that xu x
x tan
2x
1
y
2
u yy
, show that y2
yu y
3/2
2
x
z y
(sin 2y)z y
uy
z
0
x
.
, then show tat u xx .u yy
x y tan
1
u xy
2
0 2
x
; y
2
2
ux y
0
x
0 , y
0
, show that f xy
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x
2
x
y
2
y
2
.
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10.
If r 2
x a
Sol: Given r 2
2
y
b
2
, find the value of rxx + ryy
2
= ( x − a ) + ( y − b)
2
---(1)
Differentiating (1) partially with respect to x, we get 2r.rx = 2 ( x − a ) ⇒ r.rx = x − a Differentiating again partially with respect to x, we get r.rxx + rx 2 = 1
---(2)
Similarly differentiating (1) partially with respect to x ---(3) 2r.ry = 2 ( y − h ) ⇒ r.ry = y − b Differentiating (3) partially with respect to y, we get r.ryy + ry 2
---(4)
=1
Adding equation (2) and (4), we obtain 2
2
r ( rxx + ryy ) = rx + ry = 2 2
r ( rxx + ryy ) = 2 − rx − ry 2−
11.
r2 r
2
= 2−
= 2 − 1 = 1 ∴ rxx + ryy =
If z
y
f
y
x
2
x
f x
du
12.
If z
ex
y
13.
If z
xf y
14.
If u x, y
an d u
x
y
2
−
( y − b)
r yzy
y
x
x
y f
ex
yg x , show that z 1
x tan y
xy z xy
y
y x , then show that xu x
16.
Using Euler’s theorem show that xu x
17.
yu y
x
y
x y 1 2
ta n u
yu y
If u = sec
x3 y 3 then x y
x.
u x
y.
u y
0.
2
.
for the function
x y . x y 1
Where
yz y .
1
Verify Euler’s theorem for the function f x, y
1
y
.
xz x
15.
sin
x
.
2
u
= 2−
r2
(x − a)
1
g y , show that z xy
f x
sin
2
r2
, then show that xz x
y
d f
(x − a)
2 cot u
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2
2
+ (y − b)
r2
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x3 y 3 18. If u tan , show that xu x x y 1
Sol: Given that tan u = x 3 + y3
Write z =
x+y
∂z ∂x ∂z
and
∂y
sin 2u .
x 3 + y3 x+y
. Then z is a homogeneous function of degree 2 and z = tan u.
By Euler’s theorem, But
yu y
x
∂z ∂x
+ y
=
dz ∂ u ∂u 2 . = sec u du ∂ x ∂x
=
dz ∂u ∂u 2 . = sec u du ∂y ∂y
(
∂z ∂y
= 2z
---(1)
---(2)
)
(
)
Form (1) and (2), x sec2 u u x + y sec2 u u y = 2 tan u 2
i.e., xu x
+ yu y = 2 tan u.cos u
= ( 2 sin u cos u ) = sin 2u
19.
If u
log v and v x, y is a homogeneous function of degree n, then prove that
xu x
yu y
n.
Sol: Given u = log v ⇒ v = e u
Then by Euler’s theorem, x
∂v
+ y
∂x
∂v
---(1)
= nv
∂y
Now from u = log v, v = v ( x, y) We get ∂u ∂x
=
i.e.,
and
du ∂ v 1 ∂v . = . dv ∂ x v ∂x
∂v ∂x ∂u ∂y
i.e.,
∂v ∂y
∂u
= v =
---(2)
∂x
d u ∂v 1 ∂v . . = d v ∂y v ∂y
= v.
∂u
---(3)
∂y
Substituting the value of we get i.e.,
x
x .v.
∂u ∂x
∂u ∂x
+ y
+ y.v.
∂u ∂y
∂u ∂y
= nv
= n
(or) xu x + yu y = n . www.sakshieducation.com
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20.
If x x .y y .Z z
C, then prove that
Z
(1
lo g x )
x
(1
lo g Z )
Sol: Given x x .yy .ZZ = c
log (x x .y y .ZZ ) = log c x log x + y log y + X log Z = log c Differentiating partially w. r. t. x 1 x. x
+ lo g x
(1 + log Z) ∂Z ∂y
= −
1 Z Z
+
∂Z ∂x
+ lo g Z .1
∂z ∂x
= 0
= − (1 + log x )
(1 + log x ) (1 + log Z )
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