# How do you differentiate f(x)= ( x + 1 )/ ( x - 6) using the quotient rule?

$f ' \left(x\right) = \frac{- 7}{x - 6} ^ 2$

#### Explanation:

Use the formula $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} \left(u\right) - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{v} ^ 2$

Given $f \left(x\right) = \frac{x + 1}{x - 6}$

Let $u = \left(x + 1\right)$ and $v = \left(x - 6\right)$

By the formula

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} \left(u\right) - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{v} ^ 2$

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

$\frac{d}{\mathrm{dx}} \left(\frac{x + 1}{x - 6}\right) = \frac{\left(x - 6\right) \left(1 + 0\right) - \left(x + 1\right) \left(1 - 0\right)}{x - 6} ^ 2$

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

$\frac{d}{\mathrm{dx}} \left(\frac{x + 1}{x - 6}\right) = \frac{- 7}{x - 6} ^ 2$

Second solution #2, Just to check the above solution:

Simplify the given first so that

$f \left(x\right) = \frac{x + 1}{x - 6} = 1 + \frac{7}{x - 6} = 1 + 7 {\left(x - 6\right)}^{-} 1$

differentiate

$f \left(x\right) = 1 + 7 {\left(x - 6\right)}^{-} 1$

$f ' \left(x\right) = 0 + 7 \cdot \left(- 1\right) \cdot {\left(x - 6\right)}^{- 1 - 1} \cdot \frac{d}{\mathrm{dx}} \left(x - 6\right)$

$f ' \left(x\right) = - 7 {\left(x - 6\right)}^{-} 2$

$f ' \left(x\right) = \frac{- 7}{x - 6} ^ 2$

God bless....I hope the explanation is useful.