What is the derivative of 8/(\sqrt[3]{x)} ?

The derivative of 8/(\sqrt[3]{x)} is -8/(3x^{4/3)}

What is the first derivative of 8/(\sqrt[3]{x)} ?

The first derivative of 8/(\sqrt[3]{x)} is -8/(3x^{4/3)}

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derivative of

- \frac{8}{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}

= 8 \frac{d}{d x} (x^{- \frac{1}{3}})

Apply the Power Rule:

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

= 8 (- \frac{1}{3} x^{- \frac{1}{3} - 1})

Simplify

8 (- \frac{1}{3} x^{- \frac{1}{3} - 1}) : - \frac{8}{3 x ^{\frac{4}{3}}}

= - \frac{8}{3 x ^{\frac{4}{3}}}

n = 0 \sum \infty - 6 \cdot (- \frac{1}{3})^{n}

\frac{d}{d x} (x^{3} - 11 x^{2} + 31 x - 21)

\frac{d y}{d x} = \frac{y ^{2} - x ^{2}}{x y}

\int x^{3} ln (3 x) d x

\frac{d}{d x} (- 0.025 x^{2} + 8 x)

What is the derivative of 8/(\sqrt[3]{x)} ?
The derivative of 8/(\sqrt[3]{x)} is -8/(3x^{4/3)}
What is the first derivative of 8/(\sqrt[3]{x)} ?
The first derivative of 8/(\sqrt[3]{x)} is -8/(3x^{4/3)}