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 :
[c] \[y=\sqrt{\operatorname{logx}+y}\] Or \[{{y}^{2}}=\log x+y\] Or \[2y\frac{dy}{dx}=\frac{1}{x}+\frac{dy}{dx}\] or \[\frac{dy}{dx}=\frac{1}{x(2y-1)}\]You need to login to perform this action.
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