A) \[\frac{dy}{dx}=\frac{2}{1+{{x}^{2}}}\]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 :
\[y={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] \[\frac{dy}{dx}=\frac{1}{\sqrt{1-\frac{4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}}\cdot \frac{2(1+{{x}^{2}})-4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}\] 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\[|x|\,\,=1\]
You need to login to perform this action.
You will be redirected in
3 sec
