What is the derivative of 2/(\sqrt[3]{x)} ?

The derivative of 2/(\sqrt[3]{x)} is -2/(3x^{4/3)}

What is the first derivative of 2/(\sqrt[3]{x)} ?

The first derivative of 2/(\sqrt[3]{x)} is -2/(3x^{4/3)}

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derivative of

- \frac{2}{3 x ^{\frac{4}{3}}}

Solution steps

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the Power Rule:

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

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

Simplify

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

= - \frac{2}{3 x ^{\frac{4}{3}}}

f (x) = ln (x^{6} e^{x})

\int 24 + 2 x - x^{2} d x

y = \frac{6 x}{x ^{2} + 1}, (0, 0)

\int_{0}^{4} (3 x - \frac{x ^{3}}{4}) d x

What is the derivative of 2/(\sqrt[3]{x)} ?
The derivative of 2/(\sqrt[3]{x)} is -2/(3x^{4/3)}
What is the first derivative of 2/(\sqrt[3]{x)} ?
The first derivative of 2/(\sqrt[3]{x)} is -2/(3x^{4/3)}