How do you differentiate $\frac{1}{ln (x)}$ $1 / ln(x)$ ?

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How do you differentiate $\frac{1}{ln (x)}$ $1 / ln(x)$ ?

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Answer 1 · 2017-01-09T10:15:52

$\frac{d}{d x} (\frac{1}{ln x}) = - \frac{1}{x {ln (x)}^{2}}$ $d/(dx) (1/(lnx)) = -1/(xln(x)^2)$

Explanation:

Using the chain rule: if $y = ln (x)$ $y=ln(x)$ and $u = \frac{1}{y}$ $u=1/y$ then:

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

So:

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

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How do you differentiate $\frac{1}{ln (x)}$ $1 / ln(x)$ ?

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