What is the derivative of 1/(\sqrt[3]{8x^2}) ?

The derivative of 1/(\sqrt[3]{8x^2}) is -1/(3x^{5/3)}

What is the first derivative of 1/(\sqrt[3]{8x^2}) ?

The first derivative of 1/(\sqrt[3]{8x^2}) is -1/(3x^{5/3)}

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\frac{d}{d x} (\frac{1}{3 8 x ^{2}})

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

Solution steps

\frac{d}{d x} (\frac{1}{3 8 x ^{2}})

Apply exponent rule:

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

= \frac{d}{d x} ((8 x^{2})^{- \frac{1}{3}})

Apply the chain rule:

- \frac{1}{3 ( 8 x ^{2} ) ^{\frac{4}{3}}} \frac{d}{d x} (8 x^{2})

= - \frac{1}{3 ( 8 x ^{2} ) ^{\frac{4}{3}}} \frac{d}{d x} (8 x^{2})

\frac{d}{d x} (8 x^{2}) = 16 x

= - \frac{1}{3 ( 8 x ^{2} ) ^{\frac{4}{3}}} \cdot 16 x

Simplify

- \frac{1}{3 ( 8 x ^{2} ) ^{\frac{4}{3}}} \cdot 16 x : - \frac{1}{3 x ^{\frac{5}{3}}}

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

\int_{1}^{e} ln (8 x) d x

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

f (x) = \frac{3 x + 2}{x + 1}

\frac{1}{2} (6 t + 3)^{- \frac{1}{2}}

x^{2} - 7 x - 26, - x^{2} + 3 x + 2

What is the derivative of 1/(\sqrt[3]{8x^2}) ?
The derivative of 1/(\sqrt[3]{8x^2}) is -1/(3x^{5/3)}
What is the first derivative of 1/(\sqrt[3]{8x^2}) ?
The first derivative of 1/(\sqrt[3]{8x^2}) is -1/(3x^{5/3)}