# How do you use the first and second derivatives to sketch  y=(x+2)/(x-1)?

$y ' = - \frac{3}{x - 1} ^ 2$, $y ' ' = 6 \setminus {\left(x - 1\right)}^{3}$

#### Explanation:

Given: $y = \setminus \frac{x + 2}{x - 1}$

$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{\left(x - 1\right) \setminus \frac{d}{\mathrm{dx}} \left(x + 2\right) - \left(x + 2\right) \setminus \frac{d}{\mathrm{dx}} \left(x - 1\right)}{{\left(x - 1\right)}^{2}}$
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{\left(x - 1\right) \setminus \cdot 1 - \left(x + 2\right) \setminus \cdot 1}{{\left(x - 1\right)}^{2}}$
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{- 3}{{\left(x - 1\right)}^{2}}$
$\setminus \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - 3 \setminus \frac{d}{\mathrm{dx}} \setminus \frac{1}{{\left(x - 1\right)}^{2}}$
$\setminus \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - 3 \setminus \cdot \setminus \frac{- 2}{{\left(x - 1\right)}^{3}}$
$\setminus \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \setminus \frac{6}{{\left(x - 1\right)}^{3}}$