Log with trig differentiation help! (1 Viewer)

[dishab]

So the question is as follow:
If y=In(tan2x), show that dy/dx = 2sec2x cosec2x
My attempt:
=1/tan2x * 2sec^2 2x

I cant get it into the format they want. Any help would be appreciated.

 

[Drongoski]

So the question is as follow:
If y=In(tan2x), show that dy/dx = 2sec2x cosec2x
My attempt:
=1/tan2x * 2sec^2 2x

I cant get it into the format they want. Any help would be appreciated.

your answer is right unless i'm deluded.

 

[dishab]

but how do i get in the format they need

 

[shaon0]

but how do i get in the format they need

2[sec(2x)]^2/tan(2x)
=2*cos(2x)/[sin(2x)(cos(2x))^2]
=2/[sin(2x)cos(2x)]
=2csc(2x)sec(2x)