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

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

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

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

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derivative of

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

n = 0 \sum \infty 5 (4 x^{2})^{n}

\frac{d}{d x} (e^{- 0.01 x})

n = 1 \sum \infty \frac{1}{n ^{\frac{1}{n}}}

n = 0 \sum \infty (- 1)^{n + 2} \cdot t^{2 n} \cdot (\frac{1}{n !})

\int_{- 11}^{0} \frac{1}{25 - x} d x

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