The local maximum value of the function, f(x) = ((2)/(x)) ^ x ^ 2 ,...

{f}^{'} ( x ) = 0$for maximum value<br />Let$y = {\left(\dfrac{2}{x}\right)}^{{x}^{2}}$</p><p>$\ln y = {x}^{2}ln\dfrac{2}{x}$</p><p>$\dfrac{1}{y}{y}^{'} = 2xln\dfrac{2}{x} + {x}^{2}\dfrac{1}{\dfrac{2}{x}} \times \dfrac{- 2}{{x}^{2}}$</p><p>${y}^{'} = ( xy ) \left(2ln\dfrac{2}{x} - 1\right)$</p><p>${y}^{'} = {\left(\dfrac{2}{x}\right)}^{{x}^{2}}\left(x\right)\left(2ln\dfrac{2}{x} - 1\right)$</p><p>$2ln\dfrac{2}{x} = 1$</p><p>\left(\dfrac{2}{x}\right) = e\dfrac{1}{2}$

$x = 2{e}^{- \dfrac{1}{2}}$

Then maximum value will be
$f\left(2{e}^{- \dfrac{1}{2}}\right) = {\left(\dfrac{2}{2{e}^{- \dfrac{1}{2}}}\right)}^{4{e}^{- 1}} = {e}^{2{e}^{- 1}} = e\dfrac{2}{e}$