A) \[\frac{2}{\sqrt{{{x}^{2}}-1}}\]
B) \[\frac{-2}{\sqrt{{{x}^{2}}-1}}\]
C) \[\frac{1}{\left| x \right|\,\sqrt{{{x}^{2}}-1}}\]
D) \[\frac{2}{\left| x \right|\,\sqrt{{{x}^{2}}-1}}\]
Correct Answer: D
Solution :
Let \[y={{\sec }^{-1}}x+\cos e{{c}^{-1}}\frac{x}{\sqrt{{{x}^{2}}-1}}\] Put \[x=\sec \theta \Rightarrow \theta ={{\sec }^{-1}}x\] \[\therefore \,y={{\sec }^{-1}}\left( \sec \theta \right)+\cos e{{c}^{-1}}\left( \frac{\sec \theta }{\sqrt{{{\sec }^{2}}\theta -1}} \right)\] \[=\theta +{{\sin }^{-1}}\left[ \sqrt{1-{{\cos }^{2}}\theta } \right]\] \[=\theta +{{\sin }^{-1}}\left( \sin \theta \right)=\theta +\theta =2\theta =2{{\sec }^{-1}}x\] \[\frac{dy}{dx}=2\frac{d}{dx}\left( {{\sec }^{-1}}x \right)=2\times \frac{1}{\left| x \right|\sqrt{{{x}^{2}}-1}}=\frac{2}{\left| x \right|\sqrt{{{x}^{2}}-1}}\]
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