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

1 Answer
Mar 30, 2016

Answer:

#-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)#