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Answer 1 · 2017-09-06T18:21:45

$d/(dx)[x^(1/x)] = x^(1/x)[[1 - Ln x]/x^2]$

We have $y = (x)^(1/x)$

Taking natural logarithms on both sides,

$Ln y = Ln [(x)^(1/x)]$
$implies Ln y = 1/x Ln x$

Differentiating both sides,

$1/y(dy)/(dx) = -1/x^2 Ln x + 1/x 1/x$

$implies (dy)/(dx) = y [[1 - Ln x]/x^2]$

$implies (dy)/(dx) = x^(1/x)[[1 - Ln x]/x^2]$