A) \[\frac{{{y}^{2}}}{2x-2y\log x}\]
B) \[\frac{{{y}^{2}}}{2x+\log x}\]
C) \[\frac{{{y}^{2}}}{2x+2y\log x}\]
D) None of these
Correct Answer: D
Solution :
\[y={{\sqrt{x}}^{{{\sqrt{x}}^{\sqrt{x}.....\infty }}}}\Rightarrow y={{(\sqrt{x})}^{y}}\] Þ \[\log y=y\log {{x}^{1/2}}=\frac{1}{2}y\log x\] Þ \[\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left( \log x\frac{dy}{dx}+\frac{y}{x} \right)\Rightarrow \frac{dy}{dx}\left( \frac{2}{y}-\log x \right)=\frac{y}{x}\] Þ \[\frac{dy}{dx}=\frac{y.y}{x(2-y\log x)}=\frac{{{y}^{2}}}{x(2-y\log x)}\].
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