A) \[x+y\]
B) \[1+xy\]
C) 1? xy
D) \[xy-2\]
Correct Answer: B
Solution :
\[y=\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\] \[\frac{dy}{dx}=\frac{\sqrt{1-{{x}^{2}}}\frac{1}{\sqrt{1-{{x}^{2}}}}-({{\sin }^{-1}}x)\frac{1}{2}\frac{(-2x)}{\sqrt{1-{{x}^{2}}}}}{1-{{x}^{2}}}\] \[\Rightarrow (1-{{x}^{2}})\frac{dy}{dx}=1+x\left( \frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}} \right)=1+xy\].
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