# How do you differentiate y(m)=(e^mroot3(m))/(m^2+3) using the quotient rule?

Apr 13, 2017

$y ' \left(m\right) = \frac{{e}^{m} \left(3 {m}^{3} - 5 {m}^{2} + 9 m + 3\right)}{3 {m}^{\frac{2}{3}} {\left({m}^{2} + 3\right)}^{2}}$

#### Explanation:

The quotient rule states that where:

$y \left(m\right) = \frac{f \left(m\right)}{g \left(m\right)}$

then its derivative is

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

For $y \left(m\right) = \frac{{e}^{m} \sqrt[3]{m}}{{m}^{2} + 3}$ we see that:

$\left\{\begin{matrix}f \left(m\right) = {e}^{m} \sqrt[3]{m} \\ g \left(m\right) = {m}^{2} + 3\end{matrix}\right.$

To find the derivative of $f$, we will need to use the product rule. A quick review of the product rule is that the derivative of $r \left(m\right) s \left(m\right)$ is $r ' \left(m\right) s \left(m\right) + r \left(m\right) s ' \left(m\right)$.

Rewriting $f$ as $f \left(m\right) = {e}^{m} {m}^{\frac{1}{3}}$, we see that:

$f ' \left(m\right) = \left(\frac{d}{\mathrm{dm}} {e}^{m}\right) {m}^{\frac{1}{3}} + {e}^{m} \left(\frac{d}{\mathrm{dm}} {m}^{\frac{1}{3}}\right)$

$f ' \left(m\right) = {e}^{m} {m}^{\frac{1}{3}} + {e}^{m} \left(\frac{1}{3} {m}^{- \frac{2}{3}}\right)$

Simplifying by multiplying by $\frac{3 {m}^{\frac{2}{3}}}{3 {m}^{\frac{2}{3}}}$:

$f ' \left(m\right) = \frac{3 m {e}^{m} + {e}^{m}}{3 {m}^{\frac{2}{3}}} = \frac{{e}^{m} \left(3 m + 1\right)}{3 {m}^{\frac{2}{3}}}$

And:

$g ' \left(m\right) = 2 m$

Then, through the quotient rule:

$y ' \left(m\right) = \frac{\frac{{e}^{m} \left(3 m + 1\right)}{3 {m}^{\frac{2}{3}}} \left({m}^{2} + 3\right) - \left({e}^{m} {m}^{\frac{1}{3}} \left(2 m\right)\right)}{{m}^{2} + 3} ^ 2$

Again multiplying through by $3 {m}^{\frac{2}{3}}$ in the numerator and denominator:

$y ' \left(m\right) = \frac{{e}^{m} \left(3 m + 1\right) \left({m}^{2} + 3\right) - 6 {m}^{2} {e}^{m}}{3 {m}^{\frac{2}{3}} {\left({m}^{2} + 3\right)}^{2}}$

$y ' \left(m\right) = \frac{{e}^{m} \left[\left(3 m + 1\right) \left({m}^{2} + 3\right) - 6 {m}^{2}\right]}{3 {m}^{\frac{2}{3}} {\left({m}^{2} + 3\right)}^{2}}$

$y ' \left(m\right) = \frac{{e}^{m} \left(3 {m}^{3} - 5 {m}^{2} + 9 m + 3\right)}{3 {m}^{\frac{2}{3}} {\left({m}^{2} + 3\right)}^{2}}$