Explanation for differentiating $y= \frac 1{\ln x}$ The solution in my book shows 

$$\frac{dy}{dx} = \frac{\ln(x) \cdot 0 - 1 \cdot \frac{1}{x}}{\ln^2(x)}$$

Can you please explain how the differentiation is done? 

I can easily differentiate it with substitution and I can understand the denominator easily  but I have no idea how he came up with that numerator
$y=\frac{1}{\ln(x)}=(\ln(x))^{-1}.$

Now, use the power rule for differentiation and then apply the chain rule.

$y'=(-1)(\ln(x))^{-2}\frac{1}{x}=-\frac{1}{x\ln(x)^2}$.

An Alternative way would be to use the quotient rule:

$y=\frac{1}{\ln(x)}=\frac{u}{v} \implies y'=\frac{u'v-uv'}{v^2}=\frac{(0)'\ln(x)-1(\ln(x))'}{\ln(x)^2}=-\frac{1/x}{\ln(x)^2}$
y = 1/ln x

y= log e (base x)

x^y = e

taking log of both sides

y ln x =ln e

differentiate wrt x

y(1/x)+y'(ln x) = 0

y'(ln x) = -(y/x)

y' =-y/ x(ln x)

y' =-1/x(ln x)^2