Successive Differentiation
Leibnitz Rule
Solved Problems
Leibnitz’s Rule :-
Q} If y = uv y = x2excosx Find yn = ? Soln:Let u = excosx , v = x2 yn = exrn cos(x+n∈) x2 + nexrn-1 cos(x+(n-1)∈) (2x) + 𝒏(𝒏−𝟏) 𝒙 𝒏−𝟐 𝒆 𝒓 cos(x+(n-2)∈) ×2 𝟐! where r = √𝟐 ∈ = 𝐭𝐚𝐧−𝟏 (𝟏) = 𝟓𝒕⁄𝟒 Q} y =
𝒍𝒐𝒈𝒙 𝒙
P.t. y5 =
𝟓!
𝟏
𝟏
𝟏
𝒙
𝟐
𝟑
𝟒
+ + 𝟔 { 1+
𝟏
+ − logx} 𝟓
Soln:(−𝟏)𝒏−𝟏 (𝒏−𝟏)!
Let u = logx un = 𝒙𝒏 v = 𝟏⁄𝒙 y5 = u5v + 5u4v1 + 10u3v2 + 10u2v3 +5u1v4 + uv5 𝟒!
= 𝒙𝟓 × 𝟓
𝟏
𝟐𝟒
𝟏
− 𝒙
𝟓×𝟑! −𝟏 𝒙𝟒
𝟏
= 𝒙𝟔 { 𝟓 +
𝟏
𝟏
𝟏𝟎 ×𝟐!
𝒙𝟐 𝒙𝟑 −𝟏𝟐𝟎
( 𝒙𝟓 ) + 𝒍𝒐𝒈𝒙 {
𝒙 𝟓!
+
𝟏
𝒙𝟔
𝟐
}
+ 𝟑 + 𝟐 + 𝟏 − 𝒍𝒐𝒈𝒙} 𝟒
1
−𝟏
−𝟔
(𝒙𝟑 ) + 𝟏𝟎 ( 𝒙𝟐 ) ( 𝒙𝟒 ) +
Successive Differentiation
Leibnitz Rule
Solved Problems
𝒅𝒏
Q} If In = 𝒅𝒙𝒏 {𝒙𝒏 𝒍𝒐𝒈𝒙} P.t In=nIn-1 + (n-1)! Also P.t In=n! {logx + 1 + 𝟏⁄𝟐 +. . . .+ 𝟏⁄𝒏} 𝒅𝒏
Soln :- In = 𝒅𝒙𝒏 {𝒙𝒏 𝒍𝒐𝒈𝒙} 𝒅𝒏−𝟏
𝒅
= 𝒅𝒙𝒏−𝟏 {𝒅𝒙 (𝒙𝒏 . 𝒍𝒐𝒈𝒙)}
𝒅𝒏−𝟏
𝒙𝒏
= 𝒅𝒙𝒏−𝟏 { 𝒙 + (𝒍𝒐𝒈𝒙)𝒏 𝒙𝒏−𝟏 } 𝒅𝒏−𝟏
𝒅𝒏−𝟏
= 𝒅𝒙𝒏−𝟏 {𝒙 + 𝒏 𝒅𝒙𝒏−𝟏 (𝒙𝒏−𝟏 . 𝒍𝒐𝒈𝒙) In= (𝒏 − 𝟏)𝟏 + 𝒏𝑰𝒏−𝟏 𝑰𝒏
𝒏! =
𝒏−𝟏 }
𝟏
+ 𝒏
𝟏
(𝒏−𝟏)!
𝑰𝒏−𝟏
Put n= 1,2,3,…. 𝑰𝟏 𝟏 = 𝟏 + 𝑰𝟎 𝟏!
(𝒊)
𝟎!
𝑰𝒏−𝟏 (𝒏−𝟏)
𝟏
= (𝒏−𝟏) +
𝟏 (𝒏−𝟐)!
𝑰𝟐 𝟐!
𝟏
𝟏
𝟐
𝟏!
= +
(𝒊𝒊)
𝑰𝟏
𝑰𝒏−𝟐 (𝒏 − 𝟏)
Adding 𝑰𝟏
+ 𝟏!
𝑰𝟐
+ 𝟐!
𝑰𝒏−𝟐 (𝒏−𝟐)!
𝑰𝒏 𝒏
+
𝑰𝟑
+ ……+ 𝟑! 𝑰𝒏−𝟏 (𝒏−𝟏)!
𝑰𝒏−𝟏 (𝒏−𝟏)!
+
𝑰𝒏 𝒏!
=
𝑰𝟎
+ 𝟎!
𝑰𝟏 𝟏!
+
+
𝟏 𝟏 𝟏 𝟏 𝟏 + + +⋯+ + 𝟐 𝟑 𝒏−𝟏 𝒏 𝑰 𝟏 𝟏 𝟏 = 𝟎!𝟎 + 𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒏 𝟏
𝟏
𝟏
= logx +𝟏 + 𝟐 + 𝟑 + ⋯ + 𝒏
2
𝑰𝟐 𝟐!
+⋯+
Successive Differentiation
Leibnitz Rule
Solved Problems
Q) If y = 𝒙𝒏 𝒍𝒐𝒈𝒙 𝒏! 𝒚𝒏+𝟏 = 𝒙 Soln :y = 𝒙𝒏 𝒍𝒐𝒈𝒙 𝒙𝒏
𝒚𝟏 = 𝒙 + 𝒏𝒙𝒏−𝟏 𝒍𝒐𝒈𝒙 𝒙𝒚𝟏 = 𝒙𝒏 + 𝒏𝒙𝒏 𝒍𝒐𝒈𝒙 i.e. 𝒙𝒚𝟏 = 𝒙𝒏 + 𝒏𝒚 Using leibintz theorem {𝒚𝒏+𝟏 𝒙 + 𝒏𝒚𝒏 } = n! + 𝒏𝒚𝒏 𝒚𝒏+𝟏 𝒙 = 𝒏! 𝒏! 𝒚𝒏+𝟏 = 𝒙 Q) If x+y = 1 𝒅𝒏
P.t. 𝒅𝒙𝒏 (𝒙𝒏 𝒚𝒏 ) = 𝒏! {𝒚𝒏 − (nC1)2𝒚𝒏−𝟏 𝒙 + (nC2)2𝒚𝒏−𝟐 + … … + (nCn)2𝒙𝒏 (−𝟏)𝒏 } Let u = 𝒙𝒏 v = 𝒚𝒏 = (𝟏 − 𝒙)𝒏 𝒗𝟏 = 𝒏(𝟏 − 𝒙)𝒏−𝟏 (−𝟏) = 𝒏𝒚𝒏−𝟏 (−𝟏) 𝒗𝟐 = 𝒏(𝒏 − 𝟏)(𝟏 − 𝒙)𝒏−𝟐 (−𝟏)𝟐 = 𝒏(𝒏 − 𝟏)𝒚𝒏−𝟐 (−𝟏)𝟐 𝒗𝒏 = 𝒏! (−𝟏)𝒏 𝒖𝟏 = 𝒏𝒙𝒏−𝟏 𝒖𝟐 = 𝒏(𝒏 − 𝟏) 𝒙𝒏−𝟐 3
Successive Differentiation
Leibnitz Rule
Solved Problems
𝒖𝒏 = 𝒏! 𝒅𝒏
(𝒙𝒏 . 𝒚𝒏 ) = (nC0) n! 𝒚𝒏 + (nC1)n! 𝒙𝒏𝒚𝒏−𝟏 (−𝟏) + (nC2)
𝒅𝒙𝒏 𝒏! 𝟐 𝟐
𝒙 (−𝟏)𝟐 𝒏(𝒏 − 𝟏)𝒚𝒏−𝟐 + … . . + (nCn)𝒙𝒏 n!(−𝟏)𝒏
= n!{𝒚𝒏 −(nC1)2𝒚𝒏−𝟏 𝒙 +( nC2)2𝒙𝟐 𝒚𝒏−𝟐 + … … + 𝒙𝒏 (−𝟏)𝒏 (nCn)2}
Q) If 𝒚 = 𝒕𝒂𝒏𝒙 𝑷. 𝑻. 𝒚𝒏 (𝟎) − nC2𝒚𝒏−𝟐 (𝟎)+ nC4𝒚𝒏−𝟒 (𝟎) + … … = y=
𝒔𝒊𝒏𝒏𝝅 𝟐
𝒔𝒊𝒏𝒙 𝒄𝒐𝒔𝒙
ycosx = sinx By leibintz Rule 𝒚𝒏 (𝒙)𝒄𝒐𝒔𝒙 + nC1 𝒚𝒏+𝟏 (𝒙)(−𝒔𝒊𝒏𝒙) + nC2𝒚𝒏−𝟐 (𝒙)(−𝒄𝒐𝒔𝒙) + n C3𝒚𝒏−𝟑 (𝒙)(𝒔𝒊𝒏𝒙) + n
C4𝒚𝒏−𝟒 (𝒙)(𝒄𝒐𝒔𝒙) + ⋯ = 𝐬𝐢𝐧(𝒙 +
𝒏𝝅 𝟐
)
Put x = 0 (𝟎) − nC2𝒚𝒏−𝟐 (𝟎)+ nC4𝒚𝒏−𝟒 (𝟎) + … … =
4
𝒔𝒊𝒏𝒏𝝅 𝟐
Successive Differentiation
Leibnitz Rule
Solved Problems
𝟏+𝒙
Q) If y = √𝟏−𝒙 𝒚𝒏 = (𝟏 − 𝒙𝟐 )𝒚𝟏 Hence P.T (𝟏 − 𝒙𝟐 )𝒚𝒏 − (𝟐(𝒏 − 𝟏)𝒙 + 𝟏)𝒚𝒏−𝟏 − (𝒏 − 𝟏)(𝒏 − 𝟐)𝒚𝒏−𝟐 = 𝟎 Soln:- 𝒚𝟐 =
(𝟏−𝒙)𝟐 𝟐
2y𝒚𝟏 =
=
𝟏−𝒙
(𝟏−𝒙)(𝟏)−(𝟏+𝒙)(−𝟏)
2y𝒚𝟏 =
𝒚𝟏 =
𝟏+𝒙
(𝟏−𝒙)𝟐
𝟏 (𝟏−𝒙)𝟐 𝒚
=
𝒚 (𝟏−𝒙)𝟐 𝒚𝟐
𝒚 (𝟏−𝒙)𝟐 (
𝒚𝟏 =
𝟏+𝒙 ) 𝟏−𝒙
𝒚 (𝟏−𝒙)𝟐
y = (𝟏 − 𝒙𝟐 ) 𝒚𝟏 i.e (𝟏 − 𝒙𝟐 )𝒚𝟏 − 𝒚 = 𝟎 By leibnitz rule taking (n-1)th derivative = { 𝒚𝒏 (𝟏 − 𝒙𝟐 ) + (𝒏 − 𝟏)𝒚𝒏−𝟏 (−𝟐𝒙) + (𝒏−𝟏)(𝒏−𝟐) 𝟐!
𝒚𝒏−𝟏 (−𝟐) − 𝒚𝒏−𝟏 } = 𝟎
(𝟏 − 𝒙𝟐 )𝒚𝒏 − 𝟐(𝒏 + 𝒙 + 𝟏)𝒚𝒏−𝟏 − (𝒏 − 𝟏)(𝒏 − 𝟐)𝒚𝒏−𝟐 = 𝟎 5
Successive Differentiation
Leibnitz Rule
Solved Problems
Q) If y = cos(m𝐬𝐢𝐧−𝟏 𝒙) P.T. (𝟏 − 𝒙𝟐 )𝒚𝒏−𝟐 − 𝒙(𝟐𝒏 + 𝟏)𝒚𝒏−𝟏 + (𝒎𝟐 − 𝒏𝟐 )𝒚𝒏 = 𝟎 Soln:𝒚𝟏 = sin(m𝐬𝐢𝐧−𝟏 𝒙)
𝒎 √𝟏−𝒙𝟐
𝒚𝟏 (√𝟏 − 𝒙𝟐 ) = −𝒎𝒔𝒊𝒏(m𝐬𝐢𝐧−𝟏 𝒙) 𝒚𝟐 (√𝟏 − 𝒙𝟐 ) +
𝒚𝟏 (−𝟐𝒙) √𝟏−𝒙𝟐
= −m cos(m𝐬𝐢𝐧−𝟏 𝒙)
𝒎 √𝟏−𝒙𝟐
𝒚𝟐 (𝟏 − 𝒙𝟐 ) − 𝒙𝒚𝟏 = −𝒎𝟐 𝒚 𝒚𝟐 (𝟏 − 𝒙𝟐 ) − 𝒙 𝒚𝟏 + 𝒎𝟐 𝒚 = 𝟎 By leibint’s theorem {𝒚𝒏+𝟐 (𝟏 − 𝒙𝟐 ) + 𝒏𝒚𝒏+𝟏 (−𝟐𝒙) + { 𝒚𝒏+𝟏 𝒙 + 𝒏𝒚𝒏 } + 𝒎𝟐 𝒚𝒏 = 𝟎
𝒏(𝒏−𝟏) 𝟐
𝒚𝒏 (−𝟐)} −
𝒚𝒏+𝟐 (𝟏 − 𝒙𝟐 ) + 𝒚𝒏+𝟏 (−𝟐𝒏𝒙 − 𝒙) +(𝒎𝟐 − 𝒏𝟐 ) 𝒚𝒏 = 𝟎 𝒚𝒏+𝟐 (𝟏 − 𝒙𝟐 ) − (𝟐𝒏 + 𝟏)𝒙𝒚𝒏+𝟏 + (𝒎𝟐 − 𝒏𝟐 ) 𝒚𝒏 = 𝟎
Q) If 𝒚
𝟏⁄ 𝒎
+ 𝒚−
𝟏⁄ 𝒎
= 𝟐𝒙
P.T (𝒙𝟐 − 𝟏)𝒚𝒏+𝟐 + (𝟐𝒏 + 𝟏)𝒙𝒚𝒏+𝟏 +(𝒏𝟐 − 𝒎𝟐 ) 𝒚𝒏 = 𝟎 Soln:𝒚
𝟏⁄ 𝒎
+ 𝒚−
𝟏⁄ 𝒎
= 𝟐𝒙 6
Successive Differentiation
Leibnitz Rule
(𝒚
𝟏⁄ 𝟐 𝒎)
+ 𝟏 = 𝟐𝒙𝒚
(𝒚
𝟏⁄ 𝟐 𝒎)
− 𝟐𝒙. 𝒚
𝟏 𝒚 ⁄𝒎
=
𝟏⁄ 𝒎
Solved Problems
𝟏⁄ 𝒎
+ 𝟏 =0
𝟐𝒙 ± √𝟒𝒙𝟐 −𝟒 𝟐
= 𝒙 ± √𝒙𝟐 − 𝟏 y = (𝒙 ± √𝒙𝟐 − 𝟏)𝒎 𝒚𝟏 = 𝒎(𝒙 ± √𝒙𝟐 − 𝟏)𝒎−𝟏 {𝟏 + 𝒎−𝟏
𝒚𝟏 = 𝒎(𝒙 ± √𝒙𝟐 − 𝟏)
{
𝒙 √𝒙𝟐 −𝟏
}
𝒙+√𝒙𝟐 −𝟏
}
√𝒙𝟐 −𝟏 𝒎 √𝒙𝟐
𝒚𝟏 = √𝒙𝟐 − 𝟏 = 𝒎(𝒙 ± − 𝟏) 𝒚𝟐 (𝒙𝟐 − 𝟏) + 𝒙𝒚𝟏 − 𝒎𝟐 𝒚 = 𝟎 Using Leibnitz theorem {𝒚𝒏+𝟐 (𝒙𝟐 − 𝟏) + 𝒏𝒚𝒏+𝟏 (𝟐𝒙) + 𝒏𝒚𝒏 } − 𝒎𝟐 𝒚 = 𝟎
𝒏(𝒏−𝟏) 𝟐
𝒚𝒏 (𝟐)} + {𝒚𝒏+𝟏 𝒙 +
(𝒙𝟐 − 𝟏)𝒚𝒏+𝟐 + (2n+1)x𝒚𝒏+𝟏 + (𝒏𝟐 − 𝒎𝟐 )𝒚𝒏 = 𝟎 Q) If x = cos∈ ∈ = 𝟏⁄𝒎 𝒍𝒐𝒈𝒚 (1-𝒙𝟐 )𝒚𝒏+𝟐 − (𝟐𝒏 + 𝟏)𝒙𝒚𝒏+𝟏 − (𝒏𝟐 + 𝒎𝟐 )𝒚𝒏 = 𝟎 logy = m∈ y = 𝒆𝒎∈ 7
Successive Differentiation
y = 𝒆𝒎 𝐜𝐨𝐬 𝒚𝟏 = 𝒆𝒎 𝐜𝐨𝐬
Leibnitz Rule
Solved Problems
−𝟏 𝒙
−𝟏 𝒙
{
−𝒎 √𝟏−𝒙𝟐
}
𝒚𝟏 √𝟏 − 𝒙𝟐 = − 𝒎𝟐 𝒚 𝒚𝟐 (−𝟐𝒙) √𝟏−𝒙𝟐
+ 𝒚𝟐 √𝟏 − 𝒙𝟐 = − 𝒎𝒚𝟏
𝒚𝟐 (𝟏 − 𝒙𝟐 ) − 𝒙𝒚𝟏 − 𝒎𝟐 𝒚 = 𝟎
Q) If y =
𝐬𝐢𝐧−𝟏 𝒙
𝟐 P.T 𝒚 (𝟏 + 𝒙 )+(2n+3)x𝒚𝒏+𝟏 + (𝒏 + 𝒏+𝟐 𝟐
√𝟏+𝒙
𝟏)𝟐 𝒚𝒏 = 𝟎 y(√𝟏 + 𝒙𝟐 ) = 𝐥𝐨𝐠(𝒙 + √𝒙𝟐 + 𝟏) Take 2 derivatives and apply Leibnitz rule 𝒚𝟏 (√𝟏 + 𝒙𝟐 ) + 𝒚 𝟐𝒙
8
Successive Differentiation
Leibnitz Rule
Solved Problems
Q) If u = f(x) y = 𝒆𝒂𝒙 𝒖 P.T 𝑫𝒏 𝒚 = 𝒆𝒂𝒙 (𝑫 + 𝒂)𝒏 𝒖 𝒂𝒙 𝒏−𝟐 𝟐 𝒂𝒙 n 𝑫𝒏 𝒚 = nC0𝑫𝒏𝒖 𝒆𝒂𝒙 + nC1𝑫𝒏−𝟏 𝒖 (𝒒. 𝒆 ) + C2𝑫𝒖 (𝒒 . 𝒆 ) + … + nCn 𝒖 (𝒂𝒏 . 𝒆𝒂𝒙 )
= 𝒆𝒂𝒙 { nC0 𝑫𝒏 + nC1 𝑫𝒏−𝟏 𝒂 + nC2 𝑫𝒏−𝟐 𝒂+ … + nCn 𝑫𝒏−𝒏 𝒂𝒏 } 𝒖 = 𝒆𝒂𝒙 (𝑫 + 𝒂)𝒏 𝒖
(1) If y = x2 – e2x P.T. yn = 2m–2 m(n – 1) Soln.:
= x2 e2x
y
By Leibnitz's theorem, Note : Take ‘v’as that function whose derivative vanish after some derivative. = (x2e2x)m
ym = (e2x)m x2 + (e2x)m–1 (x2)1 + … = 2me2x x2 + m2 m–1 e2x
x+
m(m 1) 2
2n–2
ex
2+0
ym (0)
= 0 + 0 + m(m 1) 2
9
2m–2, e0 2
Successive Differentiation
Leibnitz Rule
Solved Problems
= 2m–2 m (n – 1)
ym (0)
(2) If y = x log (x + 1) prove that : ym = (1) Soln.:
m 2
y
(m 2 )! ( x m ) ( x 1 )m
=
x log(x 1)
By Leibnitz's theorem, = x log(x 1)
ym
m
ym
= log(x 1)
ym
=
m
ym
(1)m 1 (m 1)! m (1)m 2 (m 2)! x 1 0 (x 1)n (x 1)n1
=
ym
x m log(x 1)m 1 (x )1 ....
(1)m 2 (m 2)! (1)(m 1)x m (x 1) (x 1)m
=
10
(1)m 2 (m 2)!(x m ) (x 1)m
Successive Differentiation
Leibnitz Rule
(3) If y = xm log x P.T. ym+1 = Soln.:
Solved Problems
m! x
y = xm log x 1 x
y1 = mxm–1 logx + xm
Note : Since we want ym+1 so first find y1 and then take the mth derivative.
xy1 = m xm log x + xm
xy1 = my + xm
Now on differentiating ‘n’ times and by using Leibnitz’s theorem we have,
(x, y1)m = (ny)m + (xm)m
ym+1 x + m ym 1 = mym + m !
ym+1 =
m! x
(4) If f () = cot show that n Soln.:
f () cot
c1
f n 1 () nc 3 f n 3 () nc 5 f n 5 ().... cos
n 2
cos sin
f () sin = cos Now on differentiating ‘n’ times and by using Leibnitz’s theorem we have, f () sin cos n
n
f n () sin nc f n 1 () cos nc f n 2 ()( sin ) 1
2
11
Successive Differentiation
Leibnitz Rule
Solved Problems
n nc f n 3 ()( cos ) .. cos 2 3
put = 0, 0 nc1 f n 1 (0) 0 nc 3 f n 3 (0)... cos
n 2
n 2
nc1 f n1 () n(3 f n3 ().... cos
(5) If w = tan (logy)or y = etan–1(w) prove that : (1 + w2)yn+1 + (2nw – 1)yn n (n – 1) yn–1 = 0 y = etan–1 (w)
Soln.:
y1 = etan–1(w)
(1 + w2)y1 = y
1 1 w2
=
y 1 w2
Now on differentiating ‘n’ times and by using Leibnitz’s theorem we have,
(1 w)y1 n y n
y n 1 (1 w 2 ) n y n (2w)
1 w 2 )y n1 (2nw 1)y n n(n 1)yn 1
(6) If
x sin , y sin 2 prove
Soln.: y1 =
that (1 x
y sin 2 2 xin cos
y=
Note :
n(n 1) y n 1 (2) 0 y n 2
2x 1 x 2 2 1 x 2 2x
cos =
(2 x ) 2 1 x2
1 sin 2
1 x 2 y1 2(1 x 2 ) 2 x 2 12
2
) y n 2 (2 n 1 ) xy n 1 (n 2 4 ) y n 0
Successive Differentiation
Leibnitz Rule
Solved Problems
1 x 2 y1 2 4 x 2
on differentiating w.r.t. x, y 2 1 x 2 y1
2 8 x 2 1 x2
(1 x 2 )y 2 xy 1 8 x 1 x 2
(1 x 2 )y 2 xy 1 4 y
(1 x 2 )y 2 xy 1 4 y 0
Now on differentiating ‘n’ times by using Leibnitz’s theorem we have,
(1 x
1 x y
n 2
n(2 x )y n1
1 x y
n 2
2nxy n1 n(n 1)yn xy n1 nyn 4 yn 0
1 x y
n2
(2n 1)xy n1 [n 2 n n 4 ]yn 0
1 x y
n2
(2n 1)xy n1 (n 2 4 )yn 0
2
)y 2 n xy 1 n 4 y n 0
2
2
2
2
n(n 1) (2) yn xy n1 2ny n 4 y n 0 2
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Successive Differentiation
Leibnitz Rule
Solved Problems
(7) If y = A cos (log x) + B sin (log x)then show that x2yn+2 + (2n +1)xy constant. Soln.:
y=
n+1+
(n2 +1)n = 0 where A and B are
A cos(log x ) B sin(wgx )
y1 = A sin(log x ) B cos(log x ) x
xy1 =
x
A sin(log x ) B cos(log x )
on differentiating w.r.t. x, Aws (log x ) B sin(log x ) x x
xy2 + y1 =
x2y2 + xy1 = Aws (log x ) BSin (log x )
x2y2 + xy1 = –y … By (1)
Now on differentiating ‘n’ times by using Leibnitz’s theorem we have,
x y x, y y 2 2
n
n
x 2 y n 2 n(2 x )y n1
n
n(n 1) 2 y n1 ny n y n 2
x 2 y n2 (2n 1)xy n1 n(n 1) n 1y n 0
xy n 2 (2n 1)xy n1 (n 2 1)y n 0
(8) If y =
log( x x 2 1 )
prove that
y2m (0) = 0 and y2m+1 (0) = (–1)m, 12, 32, 52 … (2m – 1)2 Soln.:
y log(x x 2 1 ) y1
y1
2x 1 2 x x 1 2 x 1 1
2
1 x2 1 14
Successive Differentiation
Leibnitz Rule
Solved Problems
x 2 1 y1 1
on differentiating w.r.t. ‘x’ we have, x 2 1 y 2 y1
2x 2 x2 1
=0
(x 2 1)y 2 xy 1 0
Now on differentiating ‘m’ times by using Leibnitz’s theorem we have, (x
x at
1)y m 2 m (2 x )y m 1
2
2
m (m 1) 2 y m xy m 1 my m 0 2
1 ym 2 (2m 1)xy n1 m 2 y m 0
x = 0, y (0) = 0, y1(0)= 1, y2(0)= 0 ym+2(0) = –m2ym(0)
To find y3 (0)Put m = 1 in y12 (0) y 3 (0) 1 2 y1 (10 ) (1)1 1 2
m 2, y 2 2 (0) y 4 (0) 2 2 , y 2 (0) 0 m 3 y 3 2 (0) y 5 (0) 3 y 3 (0) 3 2 1 2 (1)2 1 2 3 2
= (1)
2
1 2 (2 2 1)2 )
y m(0) 0 2
and (9) If
y 2 m 1 (1)m 1 2 ,3 2....( 2m 1)2
or if p =sin–1x = sin–1y prove that (2 n 1 ) xy ( p n ) y 0 and hence de duce yn(0) =
y sin( p sin 1 x )
(1 x 2 ) y n 2
2
n 1
2
n
0 if ‘n’ is even and yn(0) = ….(32 – p2)p (n–2)2 – p2 if ‘n’ is add where ‘p’ is any constant. Soln.:
y sin( p sin 1 x )
15
Successive Differentiation
Leibnitz Rule
y1 cos( p sin 1 x )
Solved Problems
p 1 x2
1 x 2 y 1 p(p sin 1 x )
[ cos2 + sin2 =
(1 x 2 )y 2 p 2 (1 y 2 )
1] Now on differentiating ‘n’ times by using Leibnitz’s theorem we have, (1 x 2 )2 y1 y 2 y1 (2 x ) p 2 (2 yy 1 )(1 x 2 )y 2 xy 1 p 2 y 2
(1 x 2 )y n2 ny n1 (2 x )
n(n 1) y n (2) y n1 , x ny n p 2 y n 2
(1 x 2 )yn 2 (2 x 1)xy n1 (p 2 n 2 )yn 0
at x = 0,
y (o) = sin (p sin–1(o) = sin (p y1 (o) = cos (psin–1x
p 1 02
y 2 (o) p 2 y(0) 0 y n 2 (o ) (n 2 p 2 )yn(o )
If n=1,
y 3 (o) (12 p 2 )y, (o) (12 p 2 )p
n=2,
y 4 (o) (2 2 p 2 )y 2 (o) (2 2 p 2 )o 0
n=3,
y 5 (o) (3 2 p 2 )y 3 (o) (3 2 p 2 )(12 p 2 )p
16
o) = o =p
Successive Differentiation
Leibnitz Rule
Solved Problems
yn (o) o
yn (o) (n 2)2 p 2 ....( 3 2 p 2 )(12 p 2 )p
(10) If
y
p x p x
tan–1
=
( p 2 x 2 ) yn 2 2 (n 1 ) xy n 1 n(n 1 ) y n
prove
that
where ‘p’ is any constant.
p x y tan 1 p x
Soln.:
put x = p tan
y=
p p tan 1 tan tan 1 tan 1 1 tan p p tan
y=
tan 1 tan 4
y
x tan 1 ( ) 4 4 p
1 1 p 2 2 x p p x2 1 2 p
on differentiating w.r.t. ‘x’ we have,
(p2 +x2)y2 + y1 (2x)= 0
Now on differentiating ‘n’ times by wing Leibintz’s theorem we have,
(p
2
x 2 )y 2
2 xy n
1 n
0
n(n 1) 2 yn 2 x y n1 n 2 y n 0 2
( p 2 x 2 )y n 2 (2 x ) y n 2
( p 2 x 2 )y n 2 2(n 1)x y n 1 [n(n 1) 2n]y n 0
( p 2 x 2 )y n2 2(n 1)xy n1 n(n 1)y n 0
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