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

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

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

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

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Solution steps

Apply the Product Rule:

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

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

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

\frac{d}{d t} (a e^{b t})

\frac{\partial}{\partial x} (e^{- t})

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

\frac{\partial}{\partial t} (sech^{2} (x - t))

\int 6 x^{4} e^{x^{5}} d x

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