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

Answer 1

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

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)#

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Answer 2

Adam Adkins

To find the derivative of ( \csc(\frac{t}{2}) ), you can use the chain rule of differentiation.

( \frac{d}{dt} \csc(\frac{t}{2}) = -\frac{1}{\sin^2(\frac{t}{2})} \cdot \frac{d}{dt} (\frac{t}{2}) )

( = -\frac{1}{\sin^2(\frac{t}{2})} \cdot \frac{1}{2} )

( = -\frac{1}{2\sin^2(\frac{t}{2})} )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.