# What is the derivative of f(x) = (1+lnx)^2/(1-lnx)?

Oct 30, 2015

${f}^{'} \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left(1 + \ln x\right)}{x \left(1 - \ln x\right)} + \frac{1 + \ln x}{x \left(1 - \ln x\right)}$

#### Explanation:

let $u = {\left(1 + \ln x\right)}^{2}$
Let $v = \left(1 - \ln x\right)$

Note: $\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$

so $\frac{\mathrm{du}}{\mathrm{dx}} = 2 \left(1 + \ln x\right) \left(\frac{1}{x}\right) = \frac{2}{x} \left(1 + \ln x\right)$

and $\frac{\mathrm{dv}}{\mathrm{dx}} = - \frac{1}{x}$

Using $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 - \ln x\right) \times \left(\frac{2}{x}\right) \left(1 + \ln x\right)}{1 - \ln x} ^ \left(2\right) - \frac{{\left(1 + \ln x\right)}^{2} \left(- \frac{1}{x}\right)}{1 - \ln x} ^ 2$

Cancelling out:

${f}^{'} \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left(1 + \ln x\right)}{x \left(1 - \ln x\right)} + \frac{1 + \ln x}{x \left(1 - \ln x\right)}$