A) \[\frac{\sqrt{1-{{x}^{2}}}}{1+{{x}^{2}}}\]
B) \[\frac{1}{\sqrt{1-{{x}^{2}}}}\]
C) \[\frac{2}{\sqrt{1-{{x}^{2}}}(1+{{x}^{2}})}\]
D) \[\frac{2}{1+{{x}^{2}}}\]
E) \[\frac{2\sqrt{1-{{x}^{2}}}}{1+{{x}^{2}}}\]
Correct Answer: E
Solution :
Let\[u={{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)\]and\[v={{\cos }^{-1}}\sqrt{1-{{x}^{2}}}\] On differentiating w.r.t. x respectively, we get \[\frac{du}{dx}=\frac{1}{1+{{\left( \frac{2x}{1-{{x}^{2}}} \right)}^{2}}}.\left[ \frac{{{(1-x)}^{2}}2-2x(-2x)}{{{(1-{{x}^{2}})}^{2}}} \right]\] \[=\frac{2+2{{x}^{2}}}{{{(1+{{x}^{2}})}^{2}}}=\frac{2}{1+{{x}^{2}}}\] and \[\frac{dv}{dx}=-\frac{1}{\sqrt{1-(1-{{x}^{2}})}}.\left[ \frac{(-2x)}{2\sqrt{1-{{x}^{2}}}} \right]\] \[=\frac{1}{\sqrt{{{x}^{2}}}}\left[ \frac{x}{\sqrt{1-{{x}^{2}}}} \right]\] \[=\frac{1}{\sqrt{1-{{x}^{2}}}}\] \[\therefore \] \[\frac{du}{dv}=\frac{\frac{2}{1+{{x}^{2}}}}{\frac{1}{\sqrt{1-{{x}^{2}}}}}=\frac{2\sqrt{1-{{x}^{2}}}}{1+{{x}^{2}}}\]
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