# How do you find the second derivative of  f(x) = x/(1-ln(x-1)) ?

Oct 10, 2017

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

#### Explanation:

Let us write,

$f \left(x\right) = g \frac{x}{h \left(x\right)} , \text{ where, } g \left(x\right) = x , \mathmr{and} , h \left(x\right) = 1 - \ln \left(x - 1\right) .$

Using the Quotient Rule for Diffn., we have,

f'(x)={h(x)g'(x)-g(x)h'(x)}/((h(x))^2.

Now, $h \left(x\right) = 1 - \ln \left(x - 1\right) \Rightarrow h ' \left(x\right) = 0 - \frac{1}{x - 1} \cdot \frac{d}{\mathrm{dx}} \left(x - 1\right)$

$\therefore h ' \left(x\right) = - \frac{1}{x - 1} .$

Also, $g \left(x\right) = x \Rightarrow g ' \left(x\right) = 1.$

Utilising these, we get,

$f ' \left(x\right) = \frac{\left(1 - \ln \left(x - 1\right)\right) \cdot 1 - x \left(- \frac{1}{x - 1}\right)}{1 - \ln \left(x - 1\right)} ^ 2 , i . e . ,$

$f ' \left(x\right) = \frac{1 - \ln \left(x - 1\right) + \frac{x}{x - 1}}{1 - \ln \left(x - 1\right)} ^ 2 , \mathmr{and} ,$

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