What is the derivative of \sqrt[3]{1/x} ?

The derivative of \sqrt[3]{1/x} is -1/(3x^{4/3)}

What is the first derivative of \sqrt[3]{1/x} ?

The first derivative of \sqrt[3]{1/x} is -1/(3x^{4/3)}

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

Solution steps

Apply the chain rule:

\frac{1}{3 ( \frac{1}{x} ) ^{\frac{2}{3}}} \frac{d}{d x} (\frac{1}{x})

= \frac{1}{3 ( \frac{1}{x} ) ^{\frac{2}{3}}} \frac{d}{d x} (\frac{1}{x})

\frac{d}{d x} (\frac{1}{x}) = - \frac{1}{x ^{2}}

= \frac{1}{3 ( \frac{1}{x} ) ^{\frac{2}{3}}} (- \frac{1}{x ^{2}})

Simplify

\frac{1}{3 ( \frac{1}{x} ) ^{\frac{2}{3}}} (- \frac{1}{x ^{2}}) : - \frac{1}{3 x ^{\frac{4}{3}}}

= - \frac{1}{3 x ^{\frac{4}{3}}}

\int cos (20 x) cos (13 x) d x

y = - 1.97 x^{2} + 0.51 x + 8.1

\int sin (26 x) cos (17 x) d x

\int x sin (k x) d x

2 x e^{x^{2}}

What is the derivative of \sqrt[3]{1/x} ?
The derivative of \sqrt[3]{1/x} is -1/(3x^{4/3)}
What is the first derivative of \sqrt[3]{1/x} ?
The first derivative of \sqrt[3]{1/x} is -1/(3x^{4/3)}