# How do you find the derivative of  ln[x]/x^(1/3)?

Apr 14, 2017

(3-lnx)/(3x^(4/3)

#### Explanation:

Use quotient rule:

$\frac{\left(\ln \left(x\right)\right) ' \left({x}^{\frac{1}{3}}\right) - \left({x}^{\frac{1}{3}}\right) ' \left(\ln \left(x\right)\right)}{{x}^{\frac{1}{3}}} ^ 2$

Simplify:

$\frac{\left(\frac{1}{x}\right) \left({x}^{\frac{1}{3}}\right) - \left(\frac{1}{3}\right) \left({x}^{- \frac{2}{3}}\right) \left(\ln \left(x\right)\right)}{x} ^ \left(\frac{2}{3}\right)$

Rewrite $\left(\frac{1}{x}\right)$ as ${x}^{-} 1$ and merge it with ${x}^{\frac{1}{3}}$

$\frac{{x}^{- \frac{2}{3}} - \left(\frac{1}{3}\right) \left({x}^{- \frac{2}{3}} \ln x\right)}{x} ^ \left(\frac{2}{3}\right)$

Pull out a ${x}^{- \frac{2}{3}}$ from the top:

$\frac{\left({x}^{- \frac{2}{3}}\right) \left(1 - \ln \frac{x}{3}\right)}{x} ^ \left(\frac{2}{3}\right)$

Take out the negative from ${x}^{- \frac{2}{3}}$ and move it in the denominator to add exponents:

$\frac{1 - \ln \frac{x}{3}}{{x}^{\frac{4}{3}}}$

To clean it up, multiply both the top and bottom by $3$:

(3-lnx)/(3x^(4/3)