# How do you find the derivative of y=ln(-(4x^4)/(x^3-3))^5?

Oct 22, 2017

$y ' = - \frac{5 \left(12 - {x}^{3}\right)}{x \left({x}^{3} - 3\right)}$

#### Explanation:

The function in the general form:

$y = \ln \left(f \left(x\right)\right)$

then its derivative is

$y ' = \frac{1}{f} \left(x\right) \cdot f ' \left(x\right)$

where $f \left(x\right) = {\left(\textcolor{red}{g} \left(x\right)\right)}^{5}$

and its derivative is:

$f ' \left(x\right) = 5 {\left(\textcolor{red}{g} \left(x\right)\right)}^{4} \cdot g ' \left(x\right)$

where $\textcolor{red}{g \left(x\right) = - \frac{4 {x}^{4}}{{x}^{3} - 3}}$

and its derivative is:

$g ' \left(x\right) = \frac{\left(- 4 \cdot 4 {x}^{3}\right) \left({x}^{3} - 3\right) - \left(- 4 {x}^{4}\right) \left(3 {x}^{2}\right)}{{x}^{3} - 3} ^ 2$

$= \frac{- 16 {x}^{6} + 48 {x}^{3} + 12 {x}^{6}}{{x}^{3} - 3} ^ 2$

$= \frac{48 {x}^{3} - 4 {x}^{6}}{{x}^{3} - 3} ^ 2 = \frac{4 {x}^{3} \left(12 - {x}^{3}\right)}{{x}^{3} - 3} ^ 2$

Finally, it is:

$y ' = \frac{1}{\textcolor{red}{g} \left(x\right)} ^ 5 \cdot 5 {\left(\textcolor{red}{g} \left(x\right)\right)}^{4} \cdot g ' \left(x\right)$

$= \frac{1}{- \frac{4 {x}^{4}}{{x}^{3} - 3}} ^ \cancel{5} \cdot 5 \cancel{{\left(- \frac{4 {x}^{4}}{{x}^{3} - 3}\right)}^{4}} \cdot \frac{4 {x}^{3} \left(12 - {x}^{3}\right)}{{x}^{3} - 3} ^ 2$

$= - \frac{5 \cancel{\left({x}^{3} - 3\right)}}{\cancel{4} {x}^{\cancel{4}}} \cdot \frac{\cancel{4 {x}^{3}} \left(12 - {x}^{3}\right)}{{x}^{3} - 3} ^ \cancel{2}$

$= - \frac{5 \left(12 - {x}^{3}\right)}{x \left({x}^{3} - 3\right)}$