Question

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

Answer 1

$∵$ $y(x) = (x^x)^x$
$∴$ $y=x^{x^2}$
$∴$ $\frac{dy}{dx} = x^2 \cdot x^{x^2 - 1} + x ^{x^2} \ln(x) \cdot 2x$
$∴$ …\frac{dx}{dy} = \frac{1}{x^{2} + 1(1 + 2\ln x)}$$...........(i)
Now,
$\frac{d^2x}{dx^2} = \frac{d}{dx}\left((x^{x^2} + 1(1 + 2\ln x))^{-1}\right) \cdot \frac{dx}{dy}$

$=$ $\frac{-x(x^{x^2}+1(1+2\ln x))^{-2} \cdot x^{ x^2}(1+2\ln x)(x^2+2x^2\ln x+3)}{x^{x^2} \cdot (1+2\ln x)}$

$=$ $-\frac{x^{ x^2}(1+2\ln x)(x^2+3+2x^2\ln x)}{(x^{x^2} \cdot (1+2\ln x))^3}$

$\frac{d^2x}{dy^2( at \ x=1)}=−4$

$∴$ $\frac{d^2x}{dy2( at \ ^x=1)}+20=16$