First, second and third deriative of ln (tan x )

[delsoo]

Homework Statement

y=ln(tan x), dy/dx=2/(sin2x) and d2y/dx2= -4(cos 2x)/(sin 2x)^2 , show that d3y/dx3 = ((4)(3+cos 4x))/(sin 2x)^3 ... i got the solution for dy/dx and d2y/dx2 but not d3y/dx3 , can anyone show me how to get d3y/dx3 please? Thanks in advance!

Homework Equations

The Attempt at a Solution

https://www.flickr.com/photos/123101228@N03/13824984513/ (working)

 

[SammyS]

Homework Statement

y=ln(tan x), dy/dx=2/(sin2x) and d2y/dx2= -4(cos 2x)/(sin 2x)^2 , show that d3y/dx3 = ((4)(3+cos 4x))/(sin 2x)^3 ... i got the solution for dy/dx and d2y/dx2 but not d3y/dx3 , can anyone show me how to get d3y/dx3 please? Thanks in advance!

Homework Equations

The Attempt at a Solution

https://www.flickr.com/photos/123101228@N03/13824984513/ (working)

Here is an image of the photo in your link:

Of course, there are a few ways to express the functions involved.

You first derivative is correct.

Why not write $\displaystyle \ \frac{2}{\sin(2x)}\$ as $\displaystyle \ 2\csc(2x) \ ?$

That's easy to differentiate.

Then use the product rule for y''' .