Question

Find the derivative of the following function:
f(x)= $\frac{1+\frac{1}{x}}{1-\frac{1}{x}}$

Answer 1

Here $f\left(x\right)=\frac{1+\frac{1}{x}}{1-\frac{1}{x}}=\frac{x+1}{x-1}$

$\therefore f^{\prime }\left(x\right)=\frac{d}{dx}\left[\frac{x+1}{x-1}\right]$

$=\frac{(x-1)\frac{d}{dx}(x+1)-(x+1)\frac{d}{dx}(x-1)}{(x-1{)}^2}$

$=\frac{(x-1)\times 1-(x+1)\times 1}{(x-1{)}^2}$

$=\frac{x-1-x-1}{(x-1{)}^2}=\frac{-2}{(x-1{)}^2},x\ne 0,1$