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Statement and proof of leibnitz's nth derivative theorem, Schemes and Mind Maps of Engineering Mathematics

IIMT UniversityEngineering Mathematics

Very useful written in a clean and neet manner...

Typology: Schemes and Mind Maps

2022/2023

Uploaded on 12/19/2023

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Leibnitz’s Theorem:
Proof: The Proof is by the principle of mathematical induction on n.
Step 1: Take n = 1
By direct differentiation, (uv)1 = uv1 + u1v
2 n-2 2 n-1 n-1 1 n n
If u and v are functions of x possessing derivatives of the nth order,
then
(uv)n =
n C uv + n C u
0 n 1 1 n-1
v + nC u v +...+ nC u v + nC u v.
pf3
pf4
pf5

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Leibnitz’s Theorem: Proof: The Proof is by the principle of mathematical induction on n. Step 1 : Take n = 1 By direct differentiation, (uv) 1 = uv 1 + u 1 v 2 n-2 2 n-1 n-1 1 n n If u and v are functions of x possessing derivatives of the n th order, then (uv)n = n C uv + n C u 0 n 1 1 n- v + n C u v +...+ n C (^) u v + n C u v.

For n = 2, Differentiating both sides we get (uv) 2 = u 2 v+ u 1 v 1

  • u 1 v 1
  • uv 2 2 2 = u 2 v+ C 1 u 1 v 1 + C 2 uv 2 0 m 1 1 m-1 m-1 m-1 1 m m Step 2 : We assume that the theorem is true for n = m (uv)m = m C uv + m C u v + ... + m C (^) u v + m C u v. m+1 0 m+1 0 1 m 1 1 m 1 2 m- (uv) = m C u v + m C (^) u v + m C u v + m C u v + ... ... + m C u v + m C u v. m m 1 m m+

Example : If y = sin (m sin

  • x) then prove that (i) (1 – x 2 ) y 2 – xy 1 + m 2 y = 0 (ii) (1 – x^2 ) y n+
  • (2n + 1) xy n+
  • (m^2 – n^2 ) y n

1 y = cos (m sin

  • x) m

x 2 1-

y 1 x = m cos (m sin x) 2 1- (1 – x 2 ) y 1 2 = m 2 cos 2 (m sin

x) = m 2 [ 1 – sin 2 (m sin

  • 1 x)] = m^2 (1 – y^2 ).

Differentiating both sides we get (1 – x 2 )2y 1. y 2 + y 1 2 (-2x) = m 2 (- 2y. y 1 ) (1 – x^2 ) y 2

  • xy 1 + m^2 .y = 0 Applying Leibnitz’s rule we get [(1 – x 2 ) yn+2 + n c 1 (- 2x). yn+1 + n c 2 (-2) .yn ]
  • [x y n+
  • nc 1 .1. y n ] + m^2 y n

(1 – x 2 ) y n+

  • (2n + 1) xy n+ + (m 2 - n 2 ) y n