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

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

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# 1 Answer

First note:

y=sec

^{-1}(1/t) ⇔ 1/t = sec y = 1/cosy ⇔ t = cos yNow take the t-derivative of t = cos y:

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

so that

y' =-1/√(1-t²)