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

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

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

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

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\frac{1}{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{1}{3 x ^{\frac{2}{3}}} - 0

Simplify

= \frac{1}{3 x ^{\frac{2}{3}}}

\frac{d}{d x} (4 x^{2} \cdot 2^{(3 x + 1)})

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

\int 2 - x d x

x \to 3 + lim (4)

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