A) \[\frac{2xy}{2y-{{x}^{2}}}\]
B) \[\frac{xy}{y+{{x}^{2}}}\]
C) \[\frac{xy}{y-{{x}^{2}}}\]
D) \[\frac{2x}{2+\frac{{{x}^{2}}}{y}}\]
Correct Answer: A
Solution :
Since, \[y={{x}^{2}}+\frac{1}{{{x}^{2}}+\frac{1}{{{x}^{2}}\frac{1}{{{x}^{2}}+...\infty }}}\] \[\Rightarrow \] \[y={{x}^{2}}+\frac{1}{y}\] \[\Rightarrow \] \[{{y}^{2}}={{x}^{2}}y+1\] On differentiating both sides w.r.t. \[x,\]we get \[2y\frac{dy}{dx}={{x}^{2}}\frac{dy}{dx}+2xy+0\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{2xy}{2y-{{x}^{2}}}\]
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