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

$\setminus \frac{{d}^{2}}{{\mathrm{dx}}^{2}} \left(f \left(x\right)\right) = \frac{4}{x - 1} ^ 3$
Given $f \left(x\right) = \frac{x + 1}{x - 1}$
Differentiating once, $\setminus \frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{\left(x - 1\right) - \left(x + 1\right)}{x - 1} ^ 2 = \frac{- 2}{x - 1} ^ 2$
Differentiating again, we get $\setminus \frac{{d}^{2}}{{\mathrm{dx}}^{2}} \left(f \left(x\right)\right) = \frac{4}{x - 1} ^ 3$