# How do I find the derivative of lnroot3(4x-5)((3x+8)^2)?

Dec 24, 2015

Holy Jesus Christ!

#### Explanation:

First, let's identify it: it's a product which two terms will depend on chain rule. Thus, we'll need both two rules.

• Product rule states that for $y = f \left(x\right) g \left(x\right)$, then $\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

• Chain rule states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

For the first term, let's rename $u = \sqrt[3]{v}$ and $v = 4 x - 5$.

For the second term, let's rename $w = 3 x + 8$

There's no need to derivate it completely straightforward. You can do steps separately, but I do think that that might confuse things a bit.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[\left(\frac{1}{u}\right) \frac{1}{3 {v}^{\frac{2}{3}}} \left(4 x - 5\right)\right] {\left(3 x + 8\right)}^{2} + \ln \left(\sqrt[3]{4 x - 5}\right) 2 w \left(3\right)$

Substituting $u$:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[\left(\frac{1}{\sqrt[3]{v}}\right) \frac{1}{3 {v}^{\frac{2}{3}}} \left(4 x - 5\right)\right] {\left(3 x + 8\right)}^{2} + \ln \left(\sqrt[3]{4 x - 5}\right) 2 w \left(3\right)$

Substituting $v$:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[\left(\frac{1}{\sqrt[3]{4 x - 5}}\right) \frac{1}{3 {\left(4 x - 5\right)}^{\frac{2}{3}}} \left(4 x - 5\right)\right] {\left(3 x + 8\right)}^{2} + \ln \left(\sqrt[3]{4 x - 5}\right) 2 w \left(3\right)$

Substituting $w$:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[\left(\frac{1}{\sqrt[3]{4 x - 5}}\right) \frac{1}{3 {\left(4 x - 5\right)}^{\frac{2}{3}}} \left(4 x - 5\right)\right] {\left(3 x + 8\right)}^{2} + \ln \left(\sqrt[3]{4 x - 5}\right) \left(6 \left(3 x + 8\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\left(4 x - 5\right)}^{\cancel{2}}}{3 \cancel{\left(4 x - 5\right)}} {\left(3 x + 8\right)}^{2} + \ln \left(\sqrt[3]{4 x - 5}\right) \left(6 \left(3 x + 8\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(3 x + 8\right) \left[\left(\frac{\left(4 x - 5\right) \left(3 x + 8\right)}{3}\right) + 6 \ln \left(\sqrt[3]{4 x - 5}\right)\right]$

Feel free to develop it, but this is fine as a final answer!