Differential Calculus-I - New Age International
Differential Calculus-I - New Age International Differential Calculus-I - New Age International
10 A TEXTBOOK OF ENGINEERING MATHEMATICS–I \ 2 D ( e cos xsin x) ax n = 1 1 = sin x + ⋅ 2sin x cos2x 2 4 1 1 = sin x + (sin3x − sin x) 2 4 1 1 = sin x + sin3x 4 4 1 4 n ax D ( e 1 n sinx) + D ( e 4 ax sin3x) 1 ax 2 n /2 −1 = e [( a + 1) sin{ x+ ntan (1/ a)} 4 n / 2 −1 + ( a + 9) sin{ 3x + n tan ( 3/ a)}] Example Example 3: 3: 3: Find the nth derivative of the following: (i) Solution: Solution: x + 3 ( x −1)( x + 2) (i) x + 3 y = ( x −1)( x + 2) By partial fractions 2 (ii) x + 3 ( x −1)( x + 2) A B = + x − 1 x + 2 Then x + 3 = Ax ( + 2) + Bx ( −1) Taking x = 1 ⇒ A = 4/3 Equation (1), becomes x =−2 ⇒ B =−1/3 x + 3 4 1 1 1 = ⋅ − ( x −1)( x + 2) 3 ( x− 1) 3( x+ 2) 2 x ( x + 2)( 2x + 3) ⎡ x + 3 ⎤ D ⎢ ⎥ ⎣( x −1)( x + 2) ⎦ n 4 n⎛ 1 ⎞ 1 n⎛ 1 ⎞ = D D 3 ⎜ − x− 1 ⎟ 3 ⎜ x+ 2 ⎟ ⎝ ⎠ ⎝ ⎠ n n 4( −1) ⋅n! 1( −1) ⋅n! = − 3 ( x − 1) 3( x + 2) n+ 1 n+ 1 1 n ⎡ 4 1 ⎤ = ( −1) ⋅n! ⎢ − 3 n+ 1 n+ 1⎥ ⎣( x− 1) ( x+ 2) ⎦ ...(1) [Using the formula (12)]
- Page 1 and 2: 1.1 INTRODUCTION UNIT I Differentia
- Page 3 and 4: DIFFERENTIAL CALCULUS-I 3 3 d y 3 m
- Page 5 and 6: DIFFERENTIAL CALCULUS-I 5 Different
- Page 7 and 8: DIFFERENTIAL CALCULUS-I 7 dy y 1 =
- Page 9: DIFFERENTIAL CALCULUS-I 9 1 (ii) We
10 A TEXTBOOK OF ENGINEERING MATHEMATICS–I<br />
\ 2<br />
D ( e cos xsin<br />
x)<br />
ax n<br />
=<br />
1 1<br />
= sin x + ⋅ 2sin<br />
x cos2x<br />
2 4<br />
1 1<br />
= sin x + (sin3x<br />
− sin x)<br />
2 4<br />
1 1<br />
= sin x + sin3x<br />
4 4<br />
1<br />
4<br />
n<br />
ax<br />
D ( e<br />
1 n<br />
sinx)<br />
+ D ( e<br />
4<br />
ax<br />
sin3x)<br />
1 ax 2 n /2 −1<br />
= e [( a + 1) sin{ x+ ntan (1/ a)}<br />
4<br />
n / 2<br />
−1<br />
+ ( a + 9)<br />
sin{ 3x<br />
+ n tan ( 3/<br />
a)}]<br />
Example Example 3: 3: 3: Find the nth derivative of the following:<br />
(i)<br />
Solution:<br />
Solution:<br />
x + 3<br />
( x −1)(<br />
x + 2)<br />
(i)<br />
x + 3<br />
y =<br />
( x −1)(<br />
x + 2)<br />
By partial fractions<br />
2<br />
(ii)<br />
x + 3<br />
( x −1)(<br />
x + 2)<br />
A B<br />
= +<br />
x − 1 x + 2<br />
Then x + 3 = Ax ( + 2) + Bx ( −1)<br />
Taking x = 1 ⇒ A = 4/3<br />
Equation (1), becomes<br />
x =−2 ⇒ B =−1/3<br />
x + 3 4 1 1 1<br />
= ⋅ −<br />
( x −1)(<br />
x + 2)<br />
3 ( x− 1) 3( x+<br />
2)<br />
2<br />
x<br />
( x + 2)(<br />
2x<br />
+ 3)<br />
⎡ x + 3 ⎤<br />
D ⎢<br />
⎥<br />
⎣(<br />
x −1)(<br />
x + 2)<br />
⎦<br />
n 4 n⎛ 1 ⎞ 1 n⎛<br />
1 ⎞<br />
= D D<br />
3<br />
⎜ −<br />
x− 1<br />
⎟<br />
3<br />
⎜<br />
x+<br />
2<br />
⎟<br />
⎝ ⎠ ⎝ ⎠<br />
n n<br />
4( −1) ⋅n! 1( −1) ⋅n!<br />
= −<br />
3 ( x − 1) 3(<br />
x + 2)<br />
n+ 1 n+<br />
1<br />
1 n ⎡ 4 1 ⎤<br />
= ( −1) ⋅n! ⎢ −<br />
3 n+ 1 n+<br />
1⎥<br />
⎣( x− 1) ( x+<br />
2) ⎦<br />
...(1)<br />
[Using the formula (12)]