How do you find the derivative of $\frac{ln x}{x^{\frac{1}{3}}}$ $(lnx)/x^(1/3)$ ?

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How do you find the derivative of $\frac{ln x}{x^{\frac{1}{3}}}$ $(lnx)/x^(1/3)$ ?

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Answer 1 · 2016-12-19T12:46:16

$\frac{d y}{d x} = \frac{3 - ln x}{3 x^{\frac{4}{3}}}$ $dy/dx=(3-lnx)/(3x^(4/3))$

Explanation:

$y = \frac{ln x}{\sqrt[3]{x}} = \frac{ln x}{x^{\frac{1}{3}}}$ $y = lnx/root3x = lnx/x^(1/3)$

Applying the quotient rule:

$\frac{d y}{d x} = \frac{x^{\frac{1}{3}} \cdot \frac{1}{x} - ln x \cdot \frac{1}{3} x^{- \frac{2}{3}}}{x^{\frac{2}{3}}}$ $dy/dx = (x^(1/3)* 1/x - lnx* 1/3x^(-2/3))/x^(2/3)$

$= \frac{x^{- \frac{2}{3}} - \frac{1}{3} ln x \cdot x^{- \frac{2}{3}}}{x^{\frac{2}{3}}}$ $= (x^(-2/3) - 1/3lnx*x^(-2/3)) / x^(2/3)$

$= \frac{x^{- \frac{2}{3}} (1 - \frac{ln x}{3})}{x^{\frac{2}{3}}}$ $= (x^(-2/3)(1-lnx/3))/x^(2/3)$

$= \frac{1 - \frac{ln x}{3}}{x^{\frac{4}{3}}}$ $=(1-lnx/3)/x^(4/3)$

$= \frac{3 - ln x}{3 x^{\frac{4}{3}}}$ $=(3-lnx)/(3x^(4/3))$

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