What is the d/(dt)(4/3 sqrt(t^3)) ?

The d/(dt)(4/3 sqrt(t^3)) is 2sqrt(t)

What is the first d/(dt)(4/3 sqrt(t^3)) ?

The first d/(dt)(4/3 sqrt(t^3)) is 2sqrt(t)

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\frac{d}{d t} (\frac{4}{3} t^{3})

2 t

Solution steps

\frac{d}{d t} (\frac{4}{3} t^{3})

Take the constant out:

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

= \frac{4}{3} \frac{d}{d t} (t^{3})

Apply radical rule:

= \frac{4}{3} \frac{d}{d t} (t^{\frac{3}{2}})

Apply the Power Rule:

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

= \frac{4}{3} \cdot \frac{3}{2} t^{\frac{3}{2} - 1}

Simplify

\frac{4}{3} \cdot \frac{3}{2} t^{\frac{3}{2} - 1} : 2 t

= 2 t

y = x + 1, y = 0, x = 0, x = 2

\frac{6}{( s + 1 ) ^{6}}

(4 x + 7)^{3}

\int x x^{2} + 16 d x

\int - x cos (x^{2}) d x

What is the d/(dt)(4/3 sqrt(t^3)) ?
The d/(dt)(4/3 sqrt(t^3)) is 2sqrt(t)
What is the first d/(dt)(4/3 sqrt(t^3)) ?
The first d/(dt)(4/3 sqrt(t^3)) is 2sqrt(t)