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

The first d/(dt)(4t^3) is 12t^2

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d/(dt)(4t^3)

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

12 t^{2}

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule:

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

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

Simplify

= 12 t^{2}

\int_{0}^{\frac{π}{2}} cos^{5} (x) d x

2 x^{4} + \frac{3 x}{5}, at x = 0

\int (2 + 2 x) e^{(4 x + 2 x^{2})} d x

\int_{0}^{1} \frac{4 x ^{3}}{1 + x ^{4}} d x

8 y^{^{''}} + 34 y = 0, y (0) = 2, y^{^{'}} (0) = 7