Question

Differentiate the function w.r.t.x. : ${x^{x}}^{x}$

Answer 1

we have to differentiate $y=x^{x^x}$

first of all, we have to take log both sides,

$logy=log\{x^{x^x}\}\\\\logy=x^xlogx$

now let , $z=x^x$
differentiate z with respect to x,
$z'=x^x[1+logx]$

so, $logy=zlogx$

differentiate both sides with respect to x,

$\frac{1}{y}\frac{dy}{dx}=\frac{d(zlogx)}{dx}\\\\\frac{dy}{dx}=y\left[z\frac{d(logx)}{dx}+logx\frac{dz}{dx}\right]\\\\\frac{dy}{dx}=x^{x^x}\left[x^x\frac{1}{x}+z'logx\right]\\\\\frac{dy}{dx}=x^{x^x}\left[\frac{x^x}{x}+logx.x^x\{logx+1\}\right]$