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

Question

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

10/28/2013 | Talia from Los Angeles, CA

Answer 1

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

10/28/2013 | Talia from Los Angeles, CA

Answer 2

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Answer 3

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)

10/28/2013 | Andre W.