How do you find the derivative of csc (t/2)csc(t2)?

1 Answer
Mar 30, 2016

-1/2csc(t/2)cot(t/2)12csc(t2)cot(t2)

Explanation:

Note that csc(t/2)=1/sin(t/2)csc(t2)=1sin(t2)

Using chain rule:
Let u=sin(t/2)u=sin(t2) thus the original function is 1/u1u and (du)/(dt)=1/2cos(t/2)dudt=12cos(t2)
d/(du)1/u=-1/u^2=-1/(sin^2(t/2))ddu1u=1u2=1sin2(t2)
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)ddtcsc(t2)=ddu1ududt=1sin2(t2)12cos(t2)=cos(t2)2sin2(t2)=12csc(t2)cot(t2)

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)