Differential Calculus-I - New Age International

DIFFERENTIAL CALCULUS-I 9
1
(ii) We have sin ( 1 cos2
)
2
2
x = − x
Therefore,
n −x 2 1 n −x 1 n −x
x = −
D ( e
sin
1
(iii) We have cos (cos3
3cos
)
4
3
x = x + x
n
D ( e
2x
3
)
D ( e ) D ( e cos2 x)
2 2
1 −x n 1 n/2 −x −1
= e ( −1) − [5 e cos(2x+ ntan
( −2))]
2 2
1 −x n n/2
−1
= e [( −1) −5cos(2x−ntan (2))]
2
cos x)
= D ( e
4
1 n 2x
2x
3 n
cos3x)
+ D ( e
4
cosx)
1 2 2 n/2 2x −1
[(2 3 ) e cos{3x ntan
(3/ 2)}]
= + +
4
3 2 2 / 2 2
−1
e x n
+ [( 2 + 1 ) cos{ x + n tan
4
1 2x n / 2
−1
= e [ 13 cos{ 3x
+ n tan
4
n / 2
−1
+ 3(
5)
⋅ cos{ x + n tan
1
(iv) We have sin x cos2x
= (sin3x
− sin x)
2
Therefore,
n −x
1
D ( e sin x cos2x)
= D ( e
2
(v) We note that
= [(( −1)
2
1 n
sin3x)
− D ( e
2
n −x
−x
+ 3
)
( 3/
2)}
( 1/
2)}]
sin x)
1 2 2 / 2 −
−1
e x n
− [(( −1)
2
= e
2
sin{ 3x
+ n tan
1 2 2 / 2 −
−1
+ 1 ) e sin{ x + n tan
x n
1 −x
n / 2
−1
n / 2
[ 10
sin( 3x
− ntan
2 1
cos xsinx =
( 1 + cos2x)
⋅ sin x
2
3)
− 2
( 1/
2)}]
( −3)}]
( −1)}]
sin( x − nπ
/ 4)]