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

Apr 15, 2016

$\frac{6 - {x}^{2}}{{x}^{2} - x + 6} ^ 2$

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{ quotient rule }}$

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

g(x) = x $\Rightarrow g ' \left(x\right) = 1$

and h(x) $= {x}^{2} - x + 6 \Rightarrow h ' \left(x\right) = 2 x - 1$
$\text{------------------------------------------------------------------}$
substitute these values into f'(x)

$\Rightarrow f ' \left(x\right) = \frac{\left({x}^{2} - x + 6\right) .1 - x \left(2 x - 1\right)}{{x}^{2} - x + 6} ^ 2$

$= \frac{{x}^{2} - x + 6 - 2 {x}^{2} + x}{{x}^{2} - x + 6} ^ 2 = \frac{6 - {x}^{2}}{{x}^{2} - x + 6} ^ 2$