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

Given, \(y=e^{\sqrt{x \sqrt{x \sqrt{x \ldots x}}}, x>1}\)
\[
\begin{aligned}
& y=e^{\sqrt{x \sqrt{x \sqrt{x \ldots . .}}}} \\
& \log _e y=(\sqrt{x \sqrt{x \sqrt{x} \ldots . .})})^{\log _e e} \\
& \Rightarrow \quad \log y=\sqrt{x \sqrt{x \sqrt{x \ldots . .}}} \Rightarrow \log y=\sqrt{x \cdot \log y} \\
& \Rightarrow \quad(\log y)^2=x \log y \Rightarrow \frac{(\log y)^2}{\log y}=x \\
& \Rightarrow \quad \log y=x \\
&
\end{aligned}
\]
Differentiating w.r.t. \(x\), we get
\[
\frac{1}{y} \times \frac{d y}{d x}=1 \Rightarrow \frac{d y}{d x}=y
\]
Differentiating w.r.t. \(x\), we get
\[
\frac{d^2 y}{d x^2}=\frac{d y}{d x} \Rightarrow \frac{d^2 y}{d x^2}=y
\]
[from Eq. (ii) ]...(iii)
At \(x=\log _e 3\), from Eq (i), we get
\[
\log y=\log 3 \Rightarrow y=3
\]
Putting \(y=3\) into Eq. (iii), we get
\[
\frac{d^2 y}{d x^2}=3
\]