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Q. 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

KCETKCET 2022Continuity and Differentiability

Solution:

$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}} .....}} $
$e ^{ x ^{\frac{1}{2}\left[1+\frac{1}{2}+\frac{1}{4}+\ldots\right]}}= e ^{ x ^{\frac{1}{2}}}= e ^{ x ^{1}}= e ^{ x }$
$\frac{d y}{d x}=e^{x}$
$\frac{ d ^{2} y }{ dx x ^{2}}= e ^{ x } x \log _{ e }^{3}= e ^{\log _{ e }^{3}}=$