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 - \frac{1}{8} sin (2 x) d x

x \to 1 lim (lo g_{4} (x))

x \to 8 + lim (\frac{1}{x - 8} - \frac{1}{x - 8})

x \to 4 lim (\frac{5 + x}{x - 4})

n = 1 \sum \infty \frac{2}{n ^{3}}