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

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SUMMARY

The discussion focuses on finding the third derivative of the function y=ln(tan x). The first derivative is confirmed as dy/dx=2/(sin 2x), and the second derivative is d2y/dx2=-4(cos 2x)/(sin 2x)^2. Participants emphasize using the product rule for differentiation to derive the third derivative, d3y/dx3, which is expressed as ((4)(3+cos 4x))/(sin 2x)^3. The correct application of trigonometric identities and differentiation techniques is crucial for solving this problem.

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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)
 
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delsoo said:

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:

attachment.php?attachmentid=68637&stc=1&d=1397436503.jpg


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''' .
 

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