Question

Differentiate w.r.t. $x$ the function: $(logx)^{logx},x>1$

Answer 1

The correct answer is $(logx)^{log\,x}[\frac{1}{x}+\frac{log(logx)}{x}]$
Let $y=(logx)^{logx}$
Taking logarithm on both the sides,we obtain
$log\,y=log\,x.log(log\,x)$
Differentiating both sides with respect to $x$,we obtain
$\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[log\,x.xlog(log\,x)]$
$⇒\frac{1}{y}\frac{dy}{dx}=log(logx).\frac{d}{dx}(logx)+logx.\frac{d}{dx}[log(logx)]$
$⇒\frac{dy}{dx}=y[log(logx).\frac{1}{x}+logx.\frac{1}{logx}.\frac{d}{dx}(logx)]$
$⇒\frac{dy}{dx}=y[\frac{1}{x}log(log\,x)+\frac{1}{x}]$
$∴\frac{dy}{dx}=(logx)^{log\,x}[\frac{1}{x}+\frac{log(logx)}{x}]$