How do you differentiate f(x)= (3x^2+4x+2)/ (3x +1 ) using the quotient rule?

Aug 13, 2016

$f ' \left(x\right) = \frac{9 {x}^{2} + 6 x - 2}{3 x + 1} ^ 2$

Explanation:

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

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

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2}} \textcolor{w h i t e}{\frac{a}{a}} |}} \ldots \ldots . . \left(A\right)$
$\textcolor{b l u e}{\text{--------------------------------------------------------}}$

$g \left(x\right) = 3 {x}^{2} + 4 x + 2 \Rightarrow g ' \left(x\right) = 6 x + 4$

$h \left(x\right) = 3 x + 1 \Rightarrow h ' \left(x\right) = 3$
$\textcolor{b l u e}{\text{------------------------------------------------------}}$
substitute these values back into (A)

$f ' \left(x\right) = \frac{\left(3 x + 1\right) \left(6 x + 4\right) - \left(3 {x}^{2} + 4 x + 2\right) .3}{3 x + 1} ^ 2$

$= \frac{18 {x}^{2} + 18 x + 4 - 9 {x}^{2} - 12 x - 6}{3 x + 1} ^ 2$

$\Rightarrow f ' \left(x\right) = \frac{9 {x}^{2} + 6 x - 2}{3 x + 1} ^ 2$