What is the derivative of \sqrt[3]{x}(x^2+4) ?

The derivative of \sqrt[3]{x}(x^2+4) is (x^2+4+6\sqrt[3]{x}x^{5/3})/(3x^{2/3)}

What is the first derivative of \sqrt[3]{x}(x^2+4) ?

The first derivative of \sqrt[3]{x}(x^2+4) is (x^2+4+6\sqrt[3]{x}x^{5/3})/(3x^{2/3)}

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Solution steps

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

\frac{d}{d x} (x^{2} + 4) = 2 x

Simplify

What is the derivative of \sqrt[3]{x}(x^2+4) ?
The derivative of \sqrt[3]{x}(x^2+4) is (x^2+4+6\sqrt[3]{x}x^{5/3})/(3x^{2/3)}
What is the first derivative of \sqrt[3]{x}(x^2+4) ?
The first derivative of \sqrt[3]{x}(x^2+4) is (x^2+4+6\sqrt[3]{x}x^{5/3})/(3x^{2/3)}