# How do you use the product rule to differentiate (2x^2-5)^3(x^3+6x)^(4/3)?

$\frac{d}{\mathrm{dx}} \left[{\left(2 {x}^{2} - 5\right)}^{3} \cdot {\left({x}^{3} + 6 x\right)}^{\frac{4}{3}}\right]$
$= 4 {\left(2 {x}^{2} - 5\right)}^{2} {\left({x}^{3} + 6 x\right)}^{\frac{1}{3}} \left(5 {x}^{4} + 17 {x}^{2} - 10\right)$

#### Explanation:

The formula $\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{d}{\mathrm{dx}} \left(v\right) + v \frac{d}{\mathrm{dx}} \left(u\right)$

let $u = {\left(2 {x}^{2} - 5\right)}^{3}$ and $v = {\left({x}^{3} + 6 x\right)}^{\frac{4}{3}}$

Now use the formula

$\frac{d}{\mathrm{dx}} \left({\left(2 {x}^{2} - 5\right)}^{3} \cdot {\left({x}^{3} + 6 x\right)}^{\frac{4}{3}}\right) = {\left(2 {x}^{2} - 5\right)}^{3} \frac{d}{\mathrm{dx}} \left({\left({x}^{3} + 6 x\right)}^{\frac{4}{3}}\right) + \left({\left({x}^{3} + 6 x\right)}^{\frac{4}{3}}\right) \frac{d}{\mathrm{dx}} \left({\left(2 {x}^{2} - 5\right)}^{3}\right)$

$\frac{d}{\mathrm{dx}} \left({\left(2 {x}^{2} - 5\right)}^{3} \cdot {\left({x}^{3} + 6 x\right)}^{\frac{4}{3}}\right) = {\left(2 {x}^{2} - 5\right)}^{3} \cdot \left(\frac{4}{3}\right) \left({\left({x}^{3} + 6 x\right)}^{\frac{4}{3} - 1}\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{3} + 6 x\right) + \left({\left({x}^{3} + 6 x\right)}^{\frac{4}{3}}\right) \cdot 3 \left({\left(2 {x}^{2} - 5\right)}^{3 - 1}\right) \cdot \frac{d}{\mathrm{dx}} \left(2 {x}^{2} - 5\right)$

$\frac{d}{\mathrm{dx}} \left({\left(2 {x}^{2} - 5\right)}^{3} \cdot {\left({x}^{3} + 6 x\right)}^{\frac{4}{3}}\right) = {\left(2 {x}^{2} - 5\right)}^{3} \cdot \left(\frac{4}{3}\right) \left({\left({x}^{3} + 6 x\right)}^{\frac{4}{3} - 1}\right) \cdot \left(3 {x}^{2} + 6\right) + \left({\left({x}^{3} + 6 x\right)}^{\frac{4}{3}}\right) \cdot 3 \left({\left(2 {x}^{2} - 5\right)}^{3 - 1}\right) \left(4 x - 0\right)$

$\frac{d}{\mathrm{dx}} \left({\left(2 {x}^{2} - 5\right)}^{3} \cdot {\left({x}^{3} + 6 x\right)}^{\frac{4}{3}}\right)$
$= 4 {\left(2 {x}^{2} - 5\right)}^{3} \cdot \left({\left({x}^{3} + 6 x\right)}^{\frac{1}{3}}\right) \left({x}^{2} + 2\right) + 12 x {\left({x}^{3} + 6 x\right)}^{\frac{4}{3}} {\left(2 {x}^{2} - 5\right)}^{2}$

$\frac{d}{\mathrm{dx}} \left({\left(2 {x}^{2} - 5\right)}^{3} \cdot {\left({x}^{3} + 6 x\right)}^{\frac{4}{3}}\right)$
$= 4 {\left(2 {x}^{2} - 5\right)}^{2} {\left({x}^{3} + 6 x\right)}^{\frac{1}{3}} \left[\left(2 {x}^{2} - 5\right) \left({x}^{2} + 2\right) + 3 x \left({x}^{3} + 6 x\right)\right]$

$\frac{d}{\mathrm{dx}} \left({\left(2 {x}^{2} - 5\right)}^{3} \cdot {\left({x}^{3} + 6 x\right)}^{\frac{4}{3}}\right)$
$= 4 {\left(2 {x}^{2} - 5\right)}^{2} {\left({x}^{3} + 6 x\right)}^{\frac{1}{3}} \left[2 {x}^{4} - {x}^{2} - 10 + 3 {x}^{4} + 18 {x}^{2}\right]$

$\frac{d}{\mathrm{dx}} \left({\left(2 {x}^{2} - 5\right)}^{3} \cdot {\left({x}^{3} + 6 x\right)}^{\frac{4}{3}}\right)$
$= 4 {\left(2 {x}^{2} - 5\right)}^{2} {\left({x}^{3} + 6 x\right)}^{\frac{1}{3}} \left(5 {x}^{4} + 17 {x}^{2} - 10\right)$

God bless....I hope the explanation is useful.