Question

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

Answer 1

Answer: 16

$y=(x)={\left(x^x\right)}^x$
$\ln y(x)=x^2\cdot \ln x$
$\frac{1}{y(x)}\cdot y^{\prime }(x)=\frac{x^2}{x}+2x\cdot \ln x$
$y^{\prime }(x)=y(x)[x+2x\ln x]$
$y(1)=1;y^{\prime }(1)=1$
$y^{\prime \prime }(x)=y^{\prime }(x)[x+2x\cdot \ln (x)]+y(x)[1+2(1+\ln x)]$
$y^{\prime \prime }(1)=1[1+0]+1(1+2)=4$
$\frac{d^{2}y}{d{x}^2}=-{\left(\frac{dy}{dx}\right)}^3\cdot \frac{d^{2}x}{d{y}^2}$
$\Rightarrow 4=-\frac{d^{2}x}{d{y}^2}$
$\frac{d^{2}x}{dy}=-4$
$-4+20=16$