Question

$\frac{1+\log x}{1-\log x}$

Answer 1

$$\begin{gathered} \mathrm{Let}u=1+\log x;v=1-\log x \\ \mathrm{Then},u\text{'}=\frac{1}{x};v\text{'}=\frac{-1}{x} \\ \text{Using the quotient rule:} \\ \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu\text{'}-uv\text{'}}{v^2} \\ \frac{d}{dx}\left(\frac{1+\log x}{1-\log x}\right)=\frac{\left(1-\log x\right)\left({\displaystyle \frac{1}{x}}\right)-\left(1+\log x\right)\left({\displaystyle \frac{-1}{x}}\right)}{{\left(1-\log x\right)}^2} \\ =\frac{1-\log x+1+\log x}{x{\left(1-\log x\right)}^2} \\ =\frac{2}{x{\left(1-\log x\right)}^2} \end{gathered}$$