A) 0
B) x
C) \[\frac{\sqrt{1-{{y}^{2}}}}{1-2{{y}^{2}}}\]
D) \[\frac{\sqrt{1-{{y}^{2}}}}{1+2{{y}^{2}}}\]
Correct Answer: C
Solution :
\[x=y\sqrt{1-{{y}^{2}}}\] Differentiate with respect to x, \[1=\frac{dy}{dx}\sqrt{1-{{y}^{2}}}+y.\frac{1}{2\sqrt{1-{{y}^{2}}}}\,.\,(-2y)\,.\,\frac{dy}{dx}\] Þ \[1=\frac{dy}{dx}\sqrt{1-{{y}^{2}}}-\frac{{{y}^{2}}}{\sqrt{1-{{y}^{2}}}}\,.\,\frac{dy}{dx}\] Þ \[1=\frac{dy}{dx}\left[ \frac{1-{{y}^{2}}-{{y}^{2}}}{\sqrt{1-{{y}^{2}}}} \right]\]Þ \[1=\frac{dy}{dx}\left[ \frac{1-2{{y}^{2}}}{\sqrt{1-{{y}^{2}}}} \right]\] \[\frac{dy}{dx}=\frac{\sqrt{1-{{y}^{2}}}}{1-2{{y}^{2}}}\].
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