ALGEBRA AND TRIGONOMETRY, ORDINARY DIFFERENTIAL EQUATIONS ...

Paper Code: BA2007-II

Semester-II

ALGEBRA AND TRIGONOMETRY, ORDINARY DIFFERENTIAL EQUATIONS,

VECTOR CALCULUS

Bachelor of Arts (B.A.) Three Year Programme

New Scheme of Examination

DIRECTORATE OF DISTANCE EDUCATION MAHARSHI DAYANAND UNIVERSITY, ROHTAK

(A State University established under Haryana Act No. XXV of 1975) NAAC 'A+' Grade Accredited University

Copyright ? 2003, Maharshi Dayanand University, ROHTAK All Rights Reserved. No part of this publication may be reproduced or stored in a retrieval system or transmitted in any form or by any means; electronic, mechanical, photocopying, recording or otherwise, without the written

permission of the copyright holder. Maharshi Dayanand University ROHTAK ? 124 001

Sr. No. 1. 2.

3. 4. 5. 6. 7. 8. 9.

Contents Title Successive Differentiation Some General Theorems on Differentiable Functions and Expansions Curvature Singular Points Curve Tracing Reduction Formulae Rectification Quadrature Volumes and Surfaces of Solids of Revolution

Page No. 1 18

47 61 76 99 118 139 156

ankit pahal

CHAPTER ? I

SUCCESSIVE DIFFERENTIATION

1.0 STRUCTURE

1.1 Introduction 1.2 Objective 1.3 nth derivatives 1.4 Use of partial fraction 1.5 Some more nth derivatives 1.6 Leibnitz Theorem 1.7 Calculation of `nth' Derivative at x = 0

1.1 INTRODUCTION

If

y

be

a

function

of

x

i.e.

f(x).

dy

Then

= f (x) is

called

1st

derivative

of

y

w.r.t.

x

and

dx

d2y dx 2

= f (x)

is

called

2nd

derivative

of

y

w.r.t.

x

and

d3y dx3

= f (x) is

card

3rd

derivative

of

y

w.r.t.

x

and so on. In general, the nth differential co-efficients of y i.e. nth derivative of y w.r.t. x is denoted by

dny dxn .

The process of finding the differential co-efficients of a function is called successive

differentiation.

1.2 OBJECTIVE

After reading this lesson, you must be able to understand ? nth derivatives. ? nth derivatives (a) By partial fraction (b) By Leibnitz theorem. ? nthderivatives at x = 0.

Example 1 : (i) If y = cot x then find

d3y dx3

.

(ii) If y = x3 sin ax. Find

d2y dx 2

Solution : (i) Let y = cot x

dy = - cos ec2x dx

d2y dx 2

= -2cos ecx(- cos ec

x

cot

x)

= 2 cosec2 x cot x = 2(1 + cot2) cot x

= 2 (cot x + cot3x)

d3y dx3

=

2[-cosec2x

+

3

cot2x

(-cosec2

x)]

= -2(1 + 3cot2x) cosec2x

= -2(1 + 3cot2x) (1 + cot2x)

Remarks

= -2( 1 + 4cot2x + 3 cot4x)

(ii)

y = x3 sin ax

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