Differential Calculus-I - New Age International

8 A TEXTBOOK OF ENGINEERING MATHEMATICS–I
where cos 2x
2
\ 4
x
\
1
= ( 1 + cos4x)
2
1 ⎡ 1
⎤
cos = ⎢1
+ ( 1+
cos4x)
+ 2 cos2x
4
⎥
⎣ 2
⎦
=
1
4
+
1
8
1 1
+ cos4x
+ cos2x
8 2
4 3 1 1
cos x = + cos2x
+ cos4x
8 2 8
n 4 n ⎛ 3⎞
1 n 1 n
D (cos x ) = D ⎜ ⎟ + D (cos2x)
+ D (cos4x)
⎝ 8 ⎠ 2
8
n n
2 ⎛ nπ⎞ 4 ⎛ nπ⎞
= 0 + cos 2x cos 4x
2
⎜ +
2
⎟+ +
8
⎜
2
⎟
⎝ ⎠ ⎝ ⎠
n n
2 ⎛ nπ⎞ 4 ⎛ nπ⎞
= cos 2x cos 4x
2
⎜ +
2
⎟+ +
8
⎜
2
⎟
⎝ ⎠ ⎝ ⎠
Example Example 2: 2: Find the nth derivative of the following:
(i) sin h 2x
sin4x
(ii)
−x
2
e sin x
(iii) e x
(v)
Solution:
Solution:
x
(i)
1 2x
−2x
sinh
2x
sin4x
= ( e − e ) ⋅ sin4x
2
sin h2x x 3 2 cos (iv) e x x
x
e x
−
sin cos 2
ax
cos sin
2
1 2x
−2x
sin h2x sin4x
= [ e sin4x
− e sin4x]
2
n 1 n 2x n −2x
D (sinh 2x
sin4x)
= [ D ( e sin4 x) −D
( e sin4 x)]
2
2x −
−2x
e e
=
2
1 n/2 2x −1−2x −1
[20 { e sin(4x ntan2) e sin(4x ntan2)}]
=
2
+ − −
Using the formula (19)