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

The derivative of (\sqrt[3]{x}/(x^3+1)) is (x^3+1-9\sqrt[3]{x}x^{8/3})/(3x^{2/3)(x^3+1)^2}

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

The first derivative of (\sqrt[3]{x}/(x^3+1)) is (x^3+1-9\sqrt[3]{x}x^{8/3})/(3x^{2/3)(x^3+1)^2}

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Solution steps

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

Simplify

What is the derivative of (\sqrt[3]{x}/(x^3+1)) ?
The derivative of (\sqrt[3]{x}/(x^3+1)) is (x^3+1-9\sqrt[3]{x}x^{8/3})/(3x^{2/3)(x^3+1)^2}
What is the first derivative of (\sqrt[3]{x}/(x^3+1)) ?
The first derivative of (\sqrt[3]{x}/(x^3+1)) is (x^3+1-9\sqrt[3]{x}x^{8/3})/(3x^{2/3)(x^3+1)^2}