Differentiation of ln tan 2x with respect to x

[Josephine]

Differentiate with respect to x:
ln tan 2x

The answer is 4 cosec 4x

So, I've gotten
dy/dx
= 2 sec^2 2x/tan 2x
=2(1+tan^2 2x)/tan 2x
=(2/tan 2x) + 2 tan 2x
=(2 x 1/tan 2x) + 2 tan 2x
=2 cot 2x + 2/cot 2x
=2 cot^2 2x + 2
=2(cot^2 2x + 1)
=2 cosec^2 2x

How do i get 4 cosec 4x? Thank you.

 

[skeeter]

\(\displaystyle \L y = ln[tan(2x)]\)

\(\displaystyle \L y' = \frac{2sec^2(2x)}{tan(2x)}\)

\(\displaystyle \L y' = \frac{\frac{2}{cos^2(2x)}}{\frac{sin(2x)}{cos(2x)}}\)

\(\displaystyle \L y' = \frac{2}{sin(2x)cos(2x)}\)

\(\displaystyle \L y' = \frac{4}{2sin(2x)cos(2x)}\)

using the double angle identity ... \(\displaystyle \L sin(2u) = 2sin(u)cos(u)\)

\(\displaystyle \L y' = \frac{4}{sin(4x)} = 4csc(4x)\)

 

[Josephine]

Hi Skeeter! Thanks for explaining.