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

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

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

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

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Solution steps

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

\frac{d x}{d x} = 1

Simplify

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