A) \[\frac{1}{\sqrt{1-{{x}^{4}}}}\]
B) \[\frac{-1}{\sqrt{1-{{x}^{4}}}}\]
C) \[\frac{x}{\sqrt{1-{{x}^{4}}}}\]
D) \[\frac{-x}{\sqrt{1-{{x}^{4}}}}\]
Correct Answer: D
Solution :
Let\[{{x}^{2}}=\cos 2\theta \] \[y={{\tan }^{-1}}\left\{ \frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }} \right\}\] \[\Rightarrow \] \[y={{\tan }^{-1}}\left\{ \frac{\sqrt{2}\cdot \cos \theta +\sqrt{2}\sin \theta }{\sqrt{2}\cdot \cos \theta -\sqrt{2}\cdot \sin \theta } \right\}\] \[\Rightarrow \] \[y={{\tan }^{-1}}\left( \frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta } \right)\] \[\Rightarrow \] \[y={{\tan }^{-1}}\left( \frac{1+\tan \theta }{1-\tan \theta } \right)\] \[\Rightarrow \] \[y={{\tan }^{-1}}\tan \left( \frac{\pi }{4}+\theta \right)\Rightarrow y=\frac{\pi }{4}+\theta \] \[\Rightarrow \] \[y=\frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}{{x}^{2}}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1}{2}\times \frac{-1}{\sqrt{1-{{x}^{4}}}}\cdot 2x\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{-x}{\sqrt{1-{{x}^{4}}}}\]
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