If y=e^{sqrt{x sqrt{x} sqrt{x} ldots}}, x1,t | vedclass

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(A) Given,$y=e^{\sqrt{x \sqrt{x \sqrt{x \ldots}}}}, x>1$. Taking natural logarithm on both sides: $\log _e y = \sqrt{x \sqrt{x \sqrt{x \ldots}}} \log

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$3$

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(A) Given,$y=e^{\sqrt{x \sqrt{x \sqrt{x \ldots}}}}, x>1$. Taking natural logarithm on both sides: $\log _e y = \sqrt{x \sqrt{x \sqrt{x \ldots}}} \log _e e = \sqrt{x \sqrt{x \sqrt{x \ldots}}}$. Since the expression under the square root repeats,we can write: $\log _e y = \sqrt{x \log _e y}$. Squaring both sides: $(\log _e y)^2 = x \log _e y$. Dividing by $\log _e y$ (since $x>1$,$y>1$,so $\log _e y 
eq 0$): $\log _e y = x$. Now,differentiating with respect to $x$: $\frac{1}{y} \frac{d y}{d x} = 1 \implies \frac{d y}{d x} = y$. Differentiating again with respect to $x$: $\frac{d^2 y}{d x^2} = \frac{d y}{d x} = y$. At $x = \log _e 3$,we have $\log _e y = \log _e 3$,which implies $y = 3$. Therefore,$\frac{d^2 y}{d x^2} = y = 3$.

If $y=e^{\sqrt{x \sqrt{x} \sqrt{x} \ldots}}, x>1$,then $\frac{d^2 y}{d x^2}$ at $x=\log _e 3$ is

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