Finding the Second Derivative of a Trigonometric Function

[cheezeitz]

Homework Statement

y' = csc²(\vartheta / 2 )
Find y"

The Attempt at a Solution

So far, i have 2 csc (\vartheta/2) *csc(\vartheta/2)cot(\vartheta/2) * 1/2

but I'm wondering, how do you take the derivative of a half angle identity?
or does it just simplify down to csc²(\vartheta / 2) * cot(\vartheta/2)

 

[xeno_gear]

You don't need to worry about derivatives of half-angle identities here. You're right about how it simplifies, and you used the chain rule properly, except you're missing a minus sign (the derivative of csc(x) is -csc(x)cot(x)).