A) \[\frac{dy}{dx}=\frac{2x}{1+{{x}^{2}}}\] for \[|x|<1\]
B) \[\frac{dy}{dx}=\frac{-2}{1+{{x}^{2}}}\]for\[|x|>1\]
C) \[\frac{dy}{dx}=2\]for \[x=-1\]
D) \[\frac{dy}{dx}\]does not exist at\[|x|=1\]
Correct Answer: C
Solution :
\[y={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] \[\frac{dy}{dx}=\frac{1}{\sqrt{1-\frac{4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}}.\frac{2(1+{{x}^{2}})-4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}\] \[=\frac{2(1-{{x}^{2}})}{\sqrt{{{(1-{{x}^{2}})}^{2}}}}.\frac{1}{1+{{x}^{2}}}=\frac{2(1+{{x}^{2}})}{|1-{{x}^{2}}|(1+{{x}^{2}})}\] If\[|x|<1\Rightarrow {{x}^{2}}<1,\]then\[1-{{x}^{2}}>0\] \[\therefore \]\[|1-{{x}^{2}}|=1-{{x}^{2}}\Rightarrow \frac{dy}{dx}=\frac{2}{1+{{x}^{2}}}\] If \[|x|>1\Rightarrow {{x}^{2}}>1,\]then\[1-{{x}^{2}}<0\] \[\therefore \]\[|1-{{x}^{2}}|={{x}^{2}}-1\Rightarrow \frac{dy}{dx}=-\frac{2}{1+{{x}^{2}}}\] Obviously \[\frac{dy}{dx}\] does not exist for\[{{x}^{2}}=1\]or\[|x|=1\]
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