What is the derivative of \sqrt[3]{y} ?

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

What is the first derivative of \sqrt[3]{y} ?

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

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derivative of

\frac{1}{3 y ^{\frac{2}{3}}}

Solution steps

Apply radical rule:

= \frac{d}{d y} (y^{\frac{1}{3}})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= \frac{1}{3} y^{\frac{1}{3} - 1}

Simplify

\frac{1}{3} y^{\frac{1}{3} - 1} : \frac{1}{3 y ^{\frac{2}{3}}}

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

y = \frac{5}{x}

y^{^{''}} + 6 y^{^{'}} + 9 = 0

\frac{d}{d x} (e^{- 0.5 x} (4 - x) - 2)

(1 - x + x^{2}) e^{x}

\frac{\partial}{\partial x} (x^{5} - 6 x + 4 y^{3} - y^{2})

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