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
\(\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[log\,x.xlog(log\,x)]\)
\(⇒\frac{1}{y}\frac{dy}{dx}=log(logx).\frac{d}{dx}(logx)+logx.\frac{d}{dx}[log(logx)]\)
\(⇒\frac{dy}{dx}=y[log(logx).\frac{1}{x}+logx.\frac{1}{logx}.\frac{d}{dx}(logx)]\)
\(⇒\frac{dy}{dx}=y[\frac{1}{x}log(log\,x)+\frac{1}{x}]\)
\(∴\frac{dy}{dx}=(logx)^{log\,x}[\frac{1}{x}+\frac{log(logx)}{x}]\)