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Answer 1 · 2016-08-18T17:27:16

$(dy)/(dx) =-1/(xln(x))$

We have $y(u(x))$ so need to use the chain rule:

$u(x) = -1/ln(x)$

$implies (du)/(dx) = 1/(xln^2(x))$

$y = ln(u) implies (dy)/(du) = 1/u = -ln(x)$

$(dy)/(dx) = (dy)/(du)(du)/(dx)$

$(dy)/(dx) = -ln(x)*1/(xln^2(x)) = -1/(xln(x))$