Question

Differentiate :
$y=x\log x+{\left[\log x\right]}^x$

Answer 1

now split the terms

$\frac{d}{dx}\left(xlogx\right)$

$=1+logx$

$\frac{d}{dx}(logx{)}^x$

let $(logx{)}^x=t$

apply log on both sides ,$logt=xlog\left(logx\right)$

now differentiate on both sides

$\frac{t^{!}}{t}=log\left(logx\right)+\frac{1}{logx}$

$t^{!}=\left(logx{)}^x\right(log\left(logx\right)+\frac{1}{logx})$

now derivative of y is

$=1+logx+\left(logx{)}^x\right(log\left(logx\right)+\frac{1}{logx})$