# How do you differentiate y= ln (x/(x-1))?

May 12, 2016

(-1)/(x(x-1)

#### Explanation:

Using $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left(\ln f \left(x\right)\right) = \frac{1}{f \left(x\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

combined with the $\textcolor{b l u e}{\text{ chain rule}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) . g ' \left(x\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}} \textcolor{g r e e n}{\text{ A}}$
$\text{------------------------------------------------------------}$

here $f \left(g \left(x\right)\right) = \ln \left(\frac{x}{x - 1}\right) \Rightarrow f ' \left(g \left(x\right)\right) = \frac{1}{\frac{x}{x - 1}} = \frac{x - 1}{x}$

and $g \left(x\right) = \frac{x}{x - 1} \Rightarrow g ' \left(x\right) = \text{see below}$
$\text{-----------------------------------------------------------}$
To differentiate g(x) we require to use the $\textcolor{b l u e}{\text{quotient rule}}$

If f(x)$= \frac{g \left(x\right)}{h \left(x\right)} \text{ then " f'(x)=(h(x).g'(x)-g(x).h'(x))/(h(x)^2)color(green)"B}$
$\text{-----------------------------------------------------------------}$

here g(x) = x $\Rightarrow g ' \left(x\right) = 1$

and h(x) = x-1 $\Rightarrow h ' \left(x\right) = 1$
$\text{----------------------------------------------------------------}$
Substitute these values into $\textcolor{g r e e n}{\text{ B}}$

$\Rightarrow f ' \left(x\right) = \frac{\left(x - 1\right) .1 - x .1}{\left({\left(x - 1\right)}^{2}\right)} = \frac{x - 1 - x}{x - 1} ^ 2 = \frac{- 1}{x - 1} ^ 2$
$\text{------------------------------------------------------------}$
Now this is the value of g'(x) that we set out to obtain from above

Finally , 'plug' these all back into $\textcolor{g r e e n}{\text{A}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x - 1}{x} \times \frac{- 1}{x - 1} ^ 2$

$= \frac{\cancel{\left(x - 1\right)}}{x} \times \frac{- 1}{\cancel{\left(x - 1\right)} \left(x - 1\right)} = \frac{- 1}{x \left(x - 1\right)}$