What is the derivative of 2/(\sqrt[3]{x^3-9}) ?

The derivative of 2/(\sqrt[3]{x^3-9}) is -(2x^2)/((x^3-9)^{4/3)}

What is the first derivative of 2/(\sqrt[3]{x^3-9}) ?

The first derivative of 2/(\sqrt[3]{x^3-9}) is -(2x^2)/((x^3-9)^{4/3)}

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- \frac{2 x ^{2}}{( x ^{3} - 9 ) ^{\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^{3} - 9)^{- \frac{1}{3}})

Apply the chain rule:

- \frac{1}{3 ( x ^{3} - 9 ) ^{\frac{4}{3}}} \frac{d}{d x} (x^{3} - 9)

= - \frac{1}{3 ( x ^{3} - 9 ) ^{\frac{4}{3}}} \frac{d}{d x} (x^{3} - 9)

\frac{d}{d x} (x^{3} - 9) = 3 x^{2}

= 2 (- \frac{1}{3 ( x ^{3} - 9 ) ^{\frac{4}{3}}} \cdot 3 x^{2})

Simplify

2 (- \frac{1}{3 ( x ^{3} - 9 ) ^{\frac{4}{3}}} \cdot 3 x^{2}) : - \frac{2 x ^{2}}{( x ^{3} - 9 ) ^{\frac{4}{3}}}

= - \frac{2 x ^{2}}{( x ^{3} - 9 ) ^{\frac{4}{3}}}

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What is the derivative of 2/(\sqrt[3]{x^3-9}) ?
The derivative of 2/(\sqrt[3]{x^3-9}) is -(2x^2)/((x^3-9)^{4/3)}
What is the first derivative of 2/(\sqrt[3]{x^3-9}) ?
The first derivative of 2/(\sqrt[3]{x^3-9}) is -(2x^2)/((x^3-9)^{4/3)}