inverse derivative: y=sec^-1 1/t

Question

please answer, i dont understand how to go about this problem. 

Answer 1

First note:

 

y=sec^-1(1/t) ⇔ 1/t = sec y = 1/cosy ⇔ t = cos y

 

Now take the t-derivative of t = cos y:

 

1 = -(sin y) y' = -y' √(1-cos²y) = -y' √(1-t²)

 

so that

 

y' =-1/√(1-t²)