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

\frac{x ^{3} + 1 - 9 3 x x ^{\frac{8}{3}}}{3 x ^{\frac{2}{3}} ( x ^{3} + 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{\frac{1}{3 x ^{\frac{2}{3}}} ( x ^{3} + 1 ) - 3 x ^{2} 3 x}{( x ^{3} + 1 ) ^{2}} : \frac{x ^{3} + 1 - 9 3 x x ^{\frac{8}{3}}}{3 x ^{\frac{2}{3}} ( x ^{3} + 1 ) ^{2}}

= \frac{x ^{3} + 1 - 9 3 x x ^{\frac{8}{3}}}{3 x ^{\frac{2}{3}} ( x ^{3} + 1 ) ^{2}}

\int \frac{( x + 1 )}{( x - 1 ) ^{2}} d x

\int \frac{sin ^{5} ( x )}{cos ^{2} ( x )} d x

\frac{\partial}{\partial x} (6 y e^{7 x})

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

\int_{0}^{\infty} x^{4} e^{- 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}