y = ln | csc θ - cot Θ | This is easier if we do a u substitution.
Let u = csc θ - cot Θ, Then y = ln | u |
du/dΘ = - csc Θ cot Θ - (-csc2 Θ)
=. csc 2 Θ - csc Θ cot Θ = csc Θ (csc Θ - cot Θ)
Since y = ln | u |
Then dy/du = 1/u
y' = dy/dΘ = (dy/du) (du/dΘ)
y' = (1/u) ( csc 2 Θ - csc Θ cot Θ)
y' = (1/csc θ - cot Θ) ( csc Θ - cot Θ) The ( csc Θ - cot Θ) cancels. So...
y' = csc Θ
Then y'' = - csc Θ cot Θ
This problem can be done without the u substitution but the chain rule gets a little complex.