y=log(√1+cos (5x÷2)÷1-cos(5x÷2)
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Answer:
ANSWER
If y=log
1+cos3x
1−cos3x
y=log(
1+cos3x
1−cos3x
)
1/2
y=
2
1
[log(1−cos3x)−log(1+cos3x)]
Differentiate w.r. to x both sides
dx
dy
=
2
1
[
1−cos3x
1
(3sin3x)−
1+cos3x
−3sin3x
]
dx
dy
=
2
3sin3x
[
1−cos3x
1
+
1+cos3x
1
]
dx
dy
=
2
3sin3x
[
(1−cos3x)(1+cos3x)
1+cos3x+1−cos3x
]
dx
dy
=
2
3sin3x
×
1−cos
2
3x
2
dx
dy
=
1−cos
2
3x
3sin3x
dx
dy
=
sin
2
3x
3sin3x
=
sin3x
3
=3cosec 3x.
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