A) \[\frac{1}{\sqrt{1+{{x}^{2}}}}\]
B) \[-\frac{1}{2}\]
C) \[-\frac{x}{\sqrt{(1-{{x}^{4}})}}\]
D) \[\frac{x\sqrt{1+{{x}^{2}}}}{1-{{x}^{4}}}\]
Correct Answer: C
Solution :
Given,\[y={{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}}\] Put\[{{x}^{2}}=\cos 2\theta \Rightarrow \theta =\frac{{{\cos }^{-1}}{{x}^{2}}}{2}\] ? (i) \[\therefore \]\[y={{\tan }^{-1}}\left( \frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }} \right)\] \[={{\tan }^{-1}}\left( \frac{\sqrt{2}\cos \theta +\sqrt{2}\sin \theta }{\sqrt{2}\cos \theta -\sqrt{2}\sin \theta } \right)\] \[={{\tan }^{-1}}\left( \frac{1+\tan \theta }{1-\tan \theta } \right)\] \[={{\tan }^{-1}}\left[ \tan \left( \frac{\pi }{4}+\theta \right) \right]\] \[=\frac{\pi }{4}+\theta \] \[\Rightarrow \]\[y=\frac{\pi }{4}+\frac{{{\cos }^{-1}}{{x}^{2}}}{2}\] [from Eq. (i)] On differentiating w.r.t.\[x,\] we get \[\frac{dy}{dx}=0-\frac{1}{2}\frac{2x}{\sqrt{1-{{x}^{4}}}}\] \[=-\frac{x}{\sqrt{1-{{x}^{4}}}}\]
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