#### Answer

${\frac{dy}{dx}}={\frac{1}{|x|{\sqrt{(x^2-1)}}}}$

#### Work Step by Step

given that ${y}={\sec^{-1}{x}}$
so ${x}={\sec{y}}$
differentiate the abve terms with respect to y:
${\frac{dx}{dy}}={\frac{d{\sec{y}}}{dy}} ={{\sec{y}{\tan{y}}}} $
${\frac{dy}{dx}}={\frac{1}{\sec{y}\tan{y}}}$(on taking inverse of above term)
${[\frac{dy}{dx}]^2}={\frac{1}{\sec^2{y}\tan^2{y}}}$ (on squaring)
${[\frac{dy}{dx}]^2}={\frac{1}{\sec^2{y}(\sec^2{y}-1)}}$
${[\frac{dy}{dx}]^2}={\frac{1}{x^2(x^2-1)}}$
on taking square root of both sides:
${\frac{dy}{dx}}={\frac{1}{|x|{\sqrt{(x^2-1)}}}}$
thus the final answer is: ${\frac{dy}{dx}}={\frac{1}{|x|{\sqrt{(x^2-1)}}}}$