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

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Answer 1 · 2014-12-11T17:36:18

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,

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

I hope that this was helpful.

Answer 2 · 2014-12-11T17:49:00

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

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