Question

How do you find the derivative of `csc (t/2)`?

Answer 1

`-1/2csc(t/2)cot(t/2)`

Explanation:

Note that `csc(t/2)=1/sin(t/2)`

Using chain rule:
Let `u=sin(t/2)` thus the original function is `1/u` and `(du)/(dt)=1/2cos(t/2)`
`d/(du)1/u=-1/u^2=-1/(sin^2(t/2))`
Thus `d/dtcsc(t/2)=d/(du)1/u*(du)/(dt)=-1/(sin^2(t/2))*1/2cos(t/2)=-cos(t/2)/(2sin^2(t/2))=-1/2csc(t/2)cot(t/2)`

Or by quotient rule:

`(f/g)'=(f'g-g'f)/g^2`
Let `f=1`, `g=sin(t/2)`
`f'=0`,`g'=1/2cos(t/2)`
`(f/g)'=(f'g-g'f)/(g^2)=(-1/2cos(t/2))/sin^2(t/2)=-1/2csc(t/2)cot(t/2)`