What is the derivative of $lnx^lnx$ ?

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Answer 1 · 2018-05-19T22:38:05

$= 2 (ln x)/x$

$(lnx^lnx)^'$

$= (ln x lnx )^'$

$= (ln^2 x )^'$

$= 2 ln x * 1/x$

Answer 2 · 2018-05-19T22:48:03

$lnx^(lnx)*(ln(lnx)+1)/x$

$y=lnx^(lnx)=e^(ln(lnx^(lnx))$

$(y)'=(e^(ln(lnx^(lnx))))'$ $=$

$e^(ln(lnx^(lnx)))*(ln(lnx^(lnx)))'$ $=$

$lnx^(lnx)*(lnx(ln(lnx))'$ $=$

$lnx^(lnx)*(ln(lnx)/x+lnx*1/lnx(lnx)')$ $=$

$lnx^(lnx)*(ln(lnx)/x+cancel(lnx)*1/(xcancel(lnx)))$ $=$

$lnx^(lnx)*(ln(lnx)+1)/x$

What is the derivative of lnx^lnxlnx^lnx?