A) \[\frac{1}{1+{{x}^{2}}}\]
B) \[-\frac{1}{1+{{x}^{2}}}\]
C) \[-\frac{2}{1+{{x}^{2}}}\]
D) \[\frac{2}{1+{{x}^{2}}}\]
Correct Answer: D
Solution :
\[\frac{d}{dx}\left\{ {{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \right\}\] Let \[\frac{1-{{x}^{2}}}{1+{{x}^{2}}}=\cos \theta \] Þ \[1-{{x}^{2}}=(1+{{x}^{2}})\cos \theta \] Þ \[-{{x}^{2}}(1+\cos \theta )=\cos \theta -1\] Þ \[{{x}^{2}}=\frac{1-\cos \theta }{1+\cos \theta }=\frac{2{{\sin }^{2}}\frac{\theta }{2}}{2{{\cos }^{2}}\frac{\theta }{2}}={{\tan }^{2}}\frac{\theta }{2}\] or \[x=\tan \frac{\theta }{2}\] or \[\theta =2{{\tan }^{-1}}x\] So, \[\frac{d}{dx}[\theta ]=\frac{d}{dx}[2{{\tan }^{-1}}x]=\frac{2}{1+{{x}^{2}}}\].
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