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

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

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

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

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derivative of

\frac{2}{3 x ^{\frac{2}{3}}}

Solution steps

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

\frac{d}{d x} (1) = 0

= \frac{2}{3 x ^{\frac{2}{3}}} - 0

Simplify

= \frac{2}{3 x ^{\frac{2}{3}}}

\frac{d}{d x} (\frac{2}{e ^{\frac{x}{2}} + e ^{- \frac{x}{2}}})

f (x) = 5 tan (3 x)

n = 1 \sum \infty \frac{n - 5}{n + 2}

f (x) = \frac{1}{x ^{2}} - x

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

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