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

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

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

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

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derivative of

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

Solution steps

Apply radical rule:

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

Apply the Power Rule:

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

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

Simplify

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

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

\int x - (x - 2) d x

ln (w)

\frac{d}{d t} (a cos (t) + t^{2} sin (t))

g (t) = 3 t + 5 t + 9

(2 x + 8 y) d x + (8 x + 4 y) d y = 0

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