Question

What is the derivative of `sec^-1(x)`?

Answer 1

Let `y=sec^{-1}x`.

by rewriting in terms of secant,

`=> sec y=x`

by differentiating with respect to `x`,

`=> sec y tan y cdot y'=1`

by dividing by `sec y tan y`,

`=> y' = 1/{sec y tan y}`

since `sec y =x` and `tan x = sqrt{sec^2 y -1}=sqrt{x^2-1}`

`=> y'=1/{x sqrt{x^2-1}}`

Hence,

`d/dx(sec^{-1}x)=1/{x sqrt{x^2-1}}`

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I hope that this was helpful.

Answer 2

`(sec^(-1)(x))'=(1)/(xsqrt(x^(2)-1)`