A) \[\frac{{{x}^{2}}}{\sqrt{1-{{x}^{4}}}}\]
B) \[\frac{{{x}^{2}}}{\sqrt{1+{{x}^{4}}}}\]
C) \[\frac{{{x}^{2}}}{\sqrt{1+{{x}^{4}}}}\]
D) \[\frac{x}{\sqrt{1-{{x}^{4}}}}\]
Correct Answer: D
Solution :
\[y={{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}\] Put \[{{x}^{2}}=\cos 2\,\theta \] \[\therefore \] \[y={{\tan }^{-1}}\frac{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}\] \[={{\tan }^{-1}}\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }\] \[={{\tan }^{-1}}\tan \left( \frac{\pi }{4}-\theta \right)\] \[\Rightarrow \,\,y=\frac{\pi }{4}-\theta =\frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}{{x}^{2}}\] On differentia mg both sides, we get \[\frac{dy}{dx}=0-\frac{1}{2}\left( -\frac{(2x)}{\sqrt{1-{{x}^{4}}}} \right)\]
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