Question

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

• A. 3
• B. 0
• C. 5
• D. 1

Answer 1

Answer: (A)

$Y=e^{\sqrt{x\sqrt{x\sqrt{x}}}........}=e^{x^{\frac{1}{2}}{x}^{\frac{1}{4}},x^{\frac{1}{8}}.....=e^{x^{\frac{1}{2}}{x}^{\frac{1}{4}},x^{\frac{1}{8}}.....}}$
$e^{x^{\frac{1}{2}\left[1+\frac{1}{2}+\frac{1}{4}+\dots \right]}}=e^{x^{\frac{1}{2}}}=e^{x^1}=e^x$
$\frac{dy}{dx}=e^x$
$\frac{d^{2}y}{dx{x}^2}=e^{x}x{\log }_e^3=e^{{\log }_e^3}=$