Hello, delso!
I would do it this way . . .
. .
. . .
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!
here's the working
https://www.flickr.com/photos/123101...3/13824984513/