# What are the first and second derivatives of f(x)=ln((4x-5)^(1/3)/(3x+8)^2) ?

Nov 14, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4}{3 \left(4 x - 5\right)} - \frac{6}{3 x + 8}$
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{8}{3 {\left(4 x - 5\right)}^{2}} + \frac{9}{3 x + 8} ^ 2$

#### Explanation:

Use the log properties to simplify

$y = \ln \left({\left(4 x - 5\right)}^{\frac{1}{3}} / {\left(3 x + 8\right)}^{2}\right)$

$y = \ln \left({\left(4 x - 5\right)}^{\frac{1}{3}}\right) - \ln \left({\left(3 x + 8\right)}^{2}\right)$

$y = \ln \frac{4 x - 5}{3} - 2 \ln \left(3 x + 8\right)$

Say that $4 x - 5 = u$ and that $3 x + 8 = v$

$y = \ln \frac{u}{3} - 2 \ln \left(v\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \frac{d}{\mathrm{du}} \ln \left(u\right) \frac{\mathrm{du}}{\mathrm{dx}} - 2 \frac{d}{\mathrm{dv}} \ln \left(v\right) \frac{\mathrm{dv}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \cdot \frac{1}{u} \frac{d}{\mathrm{dx}} \left(4 x - 5\right) - 2 \cdot \frac{1}{v} \cdot \frac{d}{\mathrm{dx}} \left(3 x + 8\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \cdot \frac{1}{u} \cdot 4 - 2 \cdot \frac{1}{v} \cdot 3$

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

To find the second derivative, just derivate again, we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4}{3 u} - \frac{6}{v}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{4}{3} \cdot \frac{d}{\mathrm{du}} \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}} - 6 \frac{d}{\mathrm{dv}} \frac{1}{v} \frac{\mathrm{dv}}{\mathrm{dx}}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{4}{3} \cdot \left(- \frac{1}{2 {u}^{2}}\right) \cdot 4 - 6 \setminus \cdot \left(- \frac{1}{2 {v}^{2}}\right) \cdot 3$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{8}{3 {\left(4 x - 5\right)}^{2}} + \frac{9}{3 x + 8} ^ 2$