What is the derivative of 4.8t^{1/3} ?

The derivative of 4.8t^{1/3} is (1.6)/(t^{2/3)}

What is the first derivative of 4.8t^{1/3} ?

The first derivative of 4.8t^{1/3} is (1.6)/(t^{2/3)}

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derivative of

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

\frac{1.6}{t ^{\frac{2}{3}}}

Solution steps

\frac{d}{d t} (4.8 t^{\frac{1}{3}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4.8 \frac{d}{d t} (t^{\frac{1}{3}})

Apply the Power Rule:

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

= 4.8 \cdot \frac{1}{3} t^{\frac{1}{3} - 1}

Simplify

4.8 \cdot \frac{1}{3} t^{\frac{1}{3} - 1} : \frac{1.6}{t ^{\frac{2}{3}}}

= \frac{1.6}{t ^{\frac{2}{3}}}

\int_{0}^{1} x e^{9 x} d x

\frac{d}{d x} ((a^{\frac{2}{3}} - x^{\frac{2}{3}})^{\frac{3}{2}})

\frac{\partial}{\partial y} (e^{- 9 t} cos (π x))

\frac{d}{d x} ((1 - sin (x))^{3})

\int \frac{x}{e ^{- x}} d x

What is the derivative of 4.8t^{1/3} ?
The derivative of 4.8t^{1/3} is (1.6)/(t^{2/3)}
What is the first derivative of 4.8t^{1/3} ?
The first derivative of 4.8t^{1/3} is (1.6)/(t^{2/3)}