If \[y={{\sqrt{x}}^{{{\sqrt{x}}^{\sqrt{x}....\infty }}}}\], then \[\frac{dy}{dx}=\]

\[\frac{{{y}^{2}}}{2x-2y\log x}\]

\[\frac{{{y}^{2}}}{2x+\log x}\]

\[\frac{{{y}^{2}}}{2x+2y\log x}\]

If \[y={{\sqrt{x}}^{{{\sqrt{x}}^{\sqrt{x}....\infty }}}}\], then \[\fr

1.	If \[y={{\sqrt{x}}^{{{\sqrt{x}}^{\sqrt{x}....\infty }}}}\], then \[\frac{dy}{dx}=\]
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
Answer» E.

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If \[y={{\sqrt{x}}^{{{\sqrt{x}}^{\sqrt{x}....\infty }}}}\], then \[\frac{dy}{dx}=\]

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