$d/dx (x^x)?$

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Answer 1 · 2017-12-08T08:39:28

$(dy)/(dx)=x^x(1+lnx)$

we can use logarithmic differentiation

$d/(dx)(x^x)$

let

$y=x^x$

take natural logs of both sides

$lny=xlnx$

we now differentiate $wrt" "x$

the ' $LHS$ will be need the chain rule, the $RHS$ the product rule

$d/(dx)(lny)=d/(dx)(xlnx)$

$1/y(dy)/(dx)=lnxd/(dx)(x)+xd/(dx)(lnx)$

$1/y(dy)/(dx)=lnx+xxx1/x$

$(dy)/(dx)=y(1+lnx)$

$:.(dy)/(dx)=x^x(1+lnx)$

d/dx (x^x)?d/dx (x^x)?