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

x 3 x + \frac{x ^{\frac{4}{3}}}{6}

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = x^{2}, g = 3 x

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

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

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

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

Simplify

\frac{1}{2} (2 x 3 x + \frac{1}{3 x ^{\frac{2}{3}}} x^{2}) : x 3 x + \frac{x ^{\frac{4}{3}}}{6}

= x 3 x + \frac{x ^{\frac{4}{3}}}{6}

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