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