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)#