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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derivative of

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

\int \frac{2 x + 3}{x ^{2} + 3 x - 5} d x

\int ln (\frac{1}{x}) d x

\frac{d y}{d t} = (a \cdot 2) - \frac{y}{20} 2, y (0) = 0

\frac{d y}{d x} = x (4 x + 5)^{\frac{1}{3}}

f (x) = 2 x - x^{2}

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}