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

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

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)