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

\int \frac{cos ( x )}{8 - sin ( x )} d x

\int \frac{7}{9 + 16 r ^{2}} d r

\frac{d y}{d x} = 4 y + 3 e^{8 x} (x + 1), y (0) = 0

x \to 3 - lim (\frac{x + 5}{x - 3})

\int sin^{3} (x) cos^{- 4} (x) d x

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}