A) \[-\frac{y}{{{x}^{2}}+{{y}^{2}}}\]
B) \[\frac{x}{\sqrt{1-{{y}^{2}}}}\]
C) \[\frac{-y}{\sqrt{{{x}^{2}}-{{y}^{2}}}}\]
D) \[\frac{-y}{x\sqrt{{{x}^{2}}-{{y}^{2}}}}\]
Correct Answer: D
Solution :
\[u={{\sin }^{-1}}\frac{y}{x}\];\[\tan u\] \[\frac{\partial u}{\partial x}=\frac{1}{\sqrt{1-\frac{{{y}^{2}}}{{{x}^{2}}}}}.\left( -\frac{y}{{{x}^{2}}} \right)=-\frac{y}{x\sqrt{{{x}^{2}}-{{y}^{2}}}}\].
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