# How do you find the derivative of f(x)=(2x+6)/(3x^2+9)?

Mar 21, 2016

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

#### Explanation:

We should use quotient rule i.e if $f \left(x\right) = g \frac{x}{h \left(x\right)}$, then

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

As $g \left(x\right) = 2 x + 6$ and $h \left(x\right) = 3 {x}^{2} + 9$

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

$f ' \left(x\right) = \frac{6 {x}^{2} + 18 - 12 {x}^{2} - 36 x}{3 {x}^{2} + 9} ^ 2$ or

$f ' \left(x\right) = \frac{\left(- 6 {x}^{2} - 36 x + 18\right)}{3 {x}^{2} + 9} ^ 2$ or

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

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

Mar 21, 2016

I get$\text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 \left({x}^{2} + 6 x - 3\right)}{3 {\left({x}^{2} + 3\right)}^{2}}$

#### Explanation:

Quotient rule states that for $y = \frac{u}{v}$ where $u = f \left(x\right) \text{ and } v = g \left(x\right)$

Then $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

Set $y = f \left(x\right) = \frac{2 x + 6}{3 {x}^{2} + 9} = \frac{u}{v}$

Then $\frac{\mathrm{du}}{\mathrm{dx}} = 2 \text{ } \frac{\mathrm{dv}}{\mathrm{dx}} = 6 x$

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$\textcolor{b r o w n}{\implies \frac{\mathrm{dy}}{\mathrm{dx}} \textcolor{g r e e n}{= \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2} \to \frac{\left(3 {x}^{2} + 9\right) \left(2\right) - \left(2 x + 6\right) \left(6 x\right)}{{\left(3 {x}^{2} + 9\right)}^{2}}}$

Not that :${\left(3 {x}^{2} + 9\right)}^{2} = {\left(3 \left({x}^{2} + 3\right)\right)}^{2} = 9 {\left({x}^{2} + 3\right)}^{2}$

$\text{ } = \frac{2 \cancel{\left(3 {x}^{2} + 9\right)}}{{\left(3 {x}^{2} + 9\right)}^{\cancel{2}}} - \frac{2 \left(x + 3\right) \left(6 x\right)}{9 {\left({x}^{2} + 3\right)}^{2}}$

$\text{ } = \frac{2}{3 \left({x}^{3} + 3\right)} - \frac{4 x \left(x + 3\right)}{3 {\left({x}^{3} + 3\right)}^{2}}$

$\text{ } = \frac{2 \left({x}^{2} + 3\right) - 4 x \left(x + 3\right)}{3 {\left({x}^{2} + 3\right)}^{2}}$

$\text{ } = \frac{2 {x}^{2} + 6 - 4 {x}^{2} - 12 x}{3 {\left({x}^{2} + 3\right)}^{2}}$

$\text{ } = \frac{- 2 {x}^{2} - 12 x + 6}{3 {\left({x}^{2} + 3\right)}^{2}}$

$\text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 \left({x}^{2} + 6 x - 3\right)}{3 {\left({x}^{2} + 3\right)}^{2}}$

$\text{ } \textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 \left({x}^{2} + 6 x - 3\right)}{3 {\left({x}^{2} + 3\right)}^{2}}}$