Differential Calculus-I - New Age International

Differential Calculus-I - New Age International Differential Calculus-I - New Age International

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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)]

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)]

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