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

The derivative of (\sqrt[3]{x}/(1+x^2)) is (1+x^2-6\sqrt[3]{x}x^{5/3})/(3x^{2/3)(1+x^2)^2}

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

The first derivative of (\sqrt[3]{x}/(1+x^2)) is (1+x^2-6\sqrt[3]{x}x^{5/3})/(3x^{2/3)(1+x^2)^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} (1 + x^{2}) = 2 x

Simplify

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

(x - 2)^{2}

\frac{d}{d x} (sin (x) sin (2 x))

\frac{d}{d x} (3 x^{4} - 16 x^{3} - 18 x^{2})

10 x + 1

What is the derivative of (\sqrt[3]{x}/(1+x^2)) ?
The derivative of (\sqrt[3]{x}/(1+x^2)) is (1+x^2-6\sqrt[3]{x}x^{5/3})/(3x^{2/3)(1+x^2)^2}
What is the first derivative of (\sqrt[3]{x}/(1+x^2)) ?
The first derivative of (\sqrt[3]{x}/(1+x^2)) is (1+x^2-6\sqrt[3]{x}x^{5/3})/(3x^{2/3)(1+x^2)^2}