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

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

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

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

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derivative of

Solution steps

Apply the Product Rule:

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

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

Simplify

\frac{\partial}{\partial x} (x e^{y^{2}})

(2 t^{2} + 1)^{^{'}}

\frac{d}{d x} (14 e^{x})

\int arccot (3 x) d x

y^{^{''}} - 2 y^{^{'}} + y = 12 e^{3 t}

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