A) \[\frac{-{{y}^{2}}}{x\left( 2-y\,\log \,x \right)}\]
B) \[\frac{{{y}^{2}}}{2+y\,\log \,x}\]
C) \[\frac{{{y}^{2}}}{x\left( 2+y\,\log \,x \right)}\]
D) \[\frac{{{y}^{2}}}{x\left( 2-y\,\log \,x \right)}\]
Correct Answer: D
Solution :
\[y={{\left( \sqrt{x} \right)}^{{{\sqrt{x}}^{\sqrt{x}........\infty }}}}\Rightarrow y{{\left( \sqrt{x} \right)}^{y}}\] \[\Rightarrow \log y=y\left( \log \sqrt{x} \right)\] \[\Rightarrow \log y=\frac{1}{2}\left( \log \sqrt{x} \right)\] \[\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left[ y\times \frac{1}{x}+\log x\left( \frac{dy}{dc} \right) \right]\] \[\Rightarrow \frac{dy}{dx}\left\{ \frac{1}{y}-\frac{1}{2}\log x \right\}=\frac{1}{2}\frac{y}{x}\] \[\Rightarrow \frac{dy}{dx}=\frac{y}{2x}\times \frac{2y}{(2-ylogx)}=\frac{{{y}^{2}}}{x(2-ylogx)}\]You need to login to perform this action.
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