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

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