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
 A: $y=\frac{1}{\ln(x)}=(\ln(x))^{-1}.$
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:
$y=\frac{1}{\ln(x)}=\frac{u}{v} \implies y'=\frac{u'v-uv'}{v^2}=\frac{(0)'\ln(x)-1(\ln(x))'}{\ln(x)^2}=-\frac{1/x}{\ln(x)^2}$
A: Your book's solution is using the quotient rule, which says that if $f(x) = \frac{u(x)}{v(x)}$, then 
$$f'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}$$
In this case, $u(x) = 1$ and $v(x) = \ln(x)$.
A: The books solution clearly makes use of the quotient rule.  But when you differentiate a quotient with numerator a constant (like in this case, in which the numerator is equal to $1$, it can often be quicker to differentiate 
using the chain rule: 
Recall the chain rule: 

Suppose $y = g(f(x))$.  Then $y' = \dfrac{dy}{dx} = g'(f(x)) \cdot f'(x).\;$  We can use it here, too, designating $f(x) = \ln x$, and $g(x) = x^{-1}.$

And hence $\;y= g(f(x)) = (\ln x)^{-1}$. 
Using the the chain rule, we get : $$\frac{dy}{dx} = \underbrace{- \ln(x)^{-2}}_{\large g'(f(x))} \cdot \underbrace{\frac 1x}_{\large f'(x)}  =\; \color{blue}{\bf -\frac 1{(x\ln(x)^2)}}$$
A: This is quotient rule of differentiation.
$$\frac{dy}{dx}\left(\frac{f(x)}{g(x)}\right) =\frac{f'(x) \cdot g(x)-g'(x)\cdot f(x)}{\big(g(x)\big)^2}$$
A: 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
