What is the derivative of 1/(2\sqrt[3]{x^2}) ?

The derivative of 1/(2\sqrt[3]{x^2}) is -1/(3x^{5/3)}

What is the first derivative of 1/(2\sqrt[3]{x^2}) ?

The first derivative of 1/(2\sqrt[3]{x^2}) is -1/(3x^{5/3)}

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- \frac{1}{3 x ^{\frac{5}{3}}}

Solution steps

Take the constant out:

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

\frac{1}{2} (- \frac{1}{3 ( x ^{2} ) ^{\frac{4}{3}}} \cdot 2 x) : - \frac{1}{3 x ^{\frac{5}{3}}}

= - \frac{1}{3 x ^{\frac{5}{3}}}

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e^{x}, x e^{x^{2}}, [0, 1]

What is the derivative of 1/(2\sqrt[3]{x^2}) ?
The derivative of 1/(2\sqrt[3]{x^2}) is -1/(3x^{5/3)}
What is the first derivative of 1/(2\sqrt[3]{x^2}) ?
The first derivative of 1/(2\sqrt[3]{x^2}) is -1/(3x^{5/3)}