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

The derivative of 1/2 x^2\sqrt[3]{x} is x\sqrt[3]{x}+(x^{4/3})/6

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

The first derivative of 1/2 x^2\sqrt[3]{x} is x\sqrt[3]{x}+(x^{4/3})/6

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Solution steps

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

Apply the Product Rule:

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

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

Simplify

f (x) = 2 x + 3

\frac{( s + 2 )}{( s ^{2} + 9 )}

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

y = x^{3}, y = 2 x^{2}

\int 8 x^{3} - 9 x^{2} + 4 d x

What is the derivative of 1/2 x^2\sqrt[3]{x} ?
The derivative of 1/2 x^2\sqrt[3]{x} is x\sqrt[3]{x}+(x^{4/3})/6
What is the first derivative of 1/2 x^2\sqrt[3]{x} ?
The first derivative of 1/2 x^2\sqrt[3]{x} is x\sqrt[3]{x}+(x^{4/3})/6