# How do you differentiate (x-4)/(x^2+2)?

Dec 21, 2017

$f ' \left(x\right) = \frac{- {x}^{2} + 8 x + 2}{x - 4} ^ 2$

#### Explanation:

Apply the quotient rule which states:

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

Let

$g \left(x\right) = x - 4$

$h \left(x\right) = {x}^{2} + 2$

Thus,

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

$h ' \left(x\right) = 2 x$

Now plugging into the formula:

$f ' \left(x\right) = \frac{\left(1\right) \cdot \left({x}^{2} + 2\right) - \left(2 x\right) \left(x - 4\right)}{x - 4} ^ 2$

Simplify:

$f ' \left(x\right) = \frac{\left({x}^{2} + 2\right) - \left(2 {x}^{2} - 8 x\right)}{x - 4} ^ 2$

$f ' \left(x\right) = \frac{- {x}^{2} + 8 x + 2}{x - 4} ^ 2$

Dec 21, 2017

$\frac{8 x - {x}^{2} + 2}{{x}^{2} + 2} ^ 2$

#### Explanation:

$\text{differentiate using the "color(blue)"quotient rule}$

$\text{given "y=(g(x))/(h(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \textcolor{b l u e}{\text{quotient rule}}$

$g \left(x\right) = x - 4 \Rightarrow g ' \left(x\right) = 1$

$h \left(x\right) = {x}^{2} + 2 \Rightarrow h ' \left(x\right) = 2 x$

$\Rightarrow \frac{d}{\mathrm{dx}} \left(\frac{x - 4}{{x}^{2} + 2}\right)$

$= \frac{{x}^{2} + 2 - 2 x \left(x - 4\right)}{{x}^{2} + 2} ^ 2$

$= \frac{8 x - {x}^{2} + 2}{{x}^{2} + 2} ^ 2$