# How do you find the derivative of y=(x+1)/(x-1)?

Dec 23, 2016

The answer is $= = - \frac{2}{x - 1} ^ 2$

#### Explanation:

This is a ratio of polynomials.

The derivative is

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{{v}^{2}}$

Here,

$u = x + 1$, $\implies$, $u ' = 1$

$v = x - 1$, $\implies$, $v ' = 1$

So,

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

$= \frac{x - 1 - x - 1}{x - 1} ^ 2$

$= - \frac{2}{x - 1} ^ 2$

Dec 23, 2016

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

#### Explanation:

Using the quotient rule:

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

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

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

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