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Answer 1 · 2016-08-18T17:27:16

$\frac{d y}{d x} = - \frac{1}{x ln (x)}$ $(dy)/(dx) =-1/(xln(x))$

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

$u (x) = - \frac{1}{ln (x)}$ $u(x) = -1/ln(x)$

$\Rightarrow \frac{d u}{d x} = \frac{1}{x {ln}^{2} (x)}$ $implies (du)/(dx) = 1/(xln^2(x))$

$y = ln (u) \Rightarrow \frac{d y}{d u} = \frac{1}{u} = - ln (x)$ $y = ln(u) implies (dy)/(du) = 1/u = -ln(x)$

$\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ $(dy)/(dx) = (dy)/(du)(du)/(dx)$

$\frac{d y}{d x} = - ln (x) \cdot \frac{1}{x {ln}^{2} (x)} = - \frac{1}{x ln (x)}$ $(dy)/(dx) = -ln(x)*1/(xln^2(x)) = -1/(xln(x))$