If $y\left(x\right)={\left(x^x\right)}^x,x>0$ then $\frac{d^{2}x}{d{y}^2}+20$ at $x=1$ is equal to

Given,

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

Taking ${\log }_e$ both side

$\ln y\left(x\right)=x^2·\ln x$

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

$\frac{1}{y\left(x\right)}·y^{\text{'}}\left(x\right)=\frac{x^2}{x}+2x·\ln x$

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

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

Now rewriting equation (i) again we get,

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

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

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

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

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