Question

$ \frac{d}{dx}{{\cosh }^{-1}}(\sec x) $ is equal to

• A. $ sin\text{ }x $
• B. $ sec\text{ }x $
• C. $ tan\text{ }x $
• D. $ cosec\text{ }x $

Answer 1

Answer: (B)

$ \frac{d}{dx}{{\cosh }^{-1}}(\sec x) $
$ =\frac{1}{\sqrt{{{\sec }^{2}}x-1}}.\sec x\tan x $
$ \left[ \because \frac{d}{dx}{{\cosh }^{-1}}x=\frac{1}{\sqrt{{{x}^{2}}-1}} \right] $
$ =\frac{\sec x.\tan x}{\sqrt{{{\tan }^{2}}x}} $
$ =\frac{\sec x.\tan x}{\tan x} $ $ =\sec x $