If y(x) = ( x ^ x ) ^ x , x > 0 then d ^ 2 x d y ^ 2 + 20 at x = 1...

Given,

$y\left(x\right) = {\left({x}^{x}\right)}^{x}$

Taking $\log_{e}$ both side

$\ln y\left(x\right) = {x}^{2} \cdot \ln x$

Now differentiating both side w.r.t $x$ we get,

$\dfrac{1}{y\left(x\right)}$ \cdot {y}^{'}$\left(x\right) = \dfrac{{x}^{2}}{x} + 2x \cdot \ln x$

${y}^{'}\left(x\right) = y\left(x\right)\left[x + 2xlnx\right]$ .......(i)

Given $y\left(1\right) = 1$, so ${y}^{'}\left(1\right) = 1$

Now rewriting equation (i) again we get,

$\dfrac{dx}{dy} = \dfrac{1}{{x}^{{x}^{2} + 1}\left(1 + 2\ln x\right)}$

Now $\dfrac{{d}^{2}x}{d{y}^{2}} = \dfrac{d}{dx}\left({\left({x}^{{x}^{2} + 1}\left(1 + 2\ln x\right)\right)}^{- 1}\right)\dfrac{dx}{dy}$

$⇒ \dfrac{{d}^{2}x}{d{y}^{2}} = \dfrac{- {x}^{{x}^{2}}\left(1 + 2\ln x\right)\left({x}^{2} + 3 + 2{x}^{2}\ln x\right)}{{\left({x}^{{x}^{2}}\left(1 + 2\ln x\right)\right)}^{3}} \times 1$

$⇒ {\left(\dfrac{{d}^{2}x}{d{y}^{2}}\right)}_{x = 1} = - 4$

So, ${\left(\dfrac{{d}^{2}x}{d{y}^{2}}\right)}_{x = 1} + 20 = - 4 + 20 = 16$