# How do you differentiate f(x)= (7 x^2 + 3 x - 6 )/ (x- 1 ) using the quotient rule?

Jan 31, 2016

$f ' \left(x\right) = \frac{7 {x}^{2} - 14 x + 3}{x - 1} ^ 2$

#### Explanation:

The quotient rule states that for a function

$f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$

The derivative of the function is

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

Applying this to the given function, we see that

$f ' \left(x\right) = \frac{\left(x - 1\right) \frac{d}{\mathrm{dx}} \left(7 {x}^{2} + 3 x - 6\right) - \left(7 {x}^{2} + 3 x - 6\right) \frac{d}{\mathrm{dx}} \left(x + 1\right)}{x + 1} ^ 2$

Both of these derivatives can be found through the power rule.

$f ' \left(x\right) = \frac{\left(x - 1\right) \left(14 x + 3\right) - \left(7 {x}^{2} + 3 x - 6\right) \left(1\right)}{x - 1} ^ 2$

Distribute and simplify.

$f ' \left(x\right) = \frac{14 {x}^{2} - 11 x - 3 - 7 {x}^{2} - 3 x + 6}{x - 1} ^ 2$

$f ' \left(x\right) = \frac{7 {x}^{2} - 14 x + 3}{x - 1} ^ 2$