Question

Log√1-cos3x/√1+cos3x differentiate with respect to x

Answer 1

Step-by-step explanation:

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.