What is the first d/(dt)(8t^3) ?

The first d/(dt)(8t^3) is 24t^2

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d/(dt)(8t^3)

\frac{d}{d t} (8 t^{3})

24 t^{2}

Solution steps

\frac{d}{d t} (8 t^{3})

Take the constant out:

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

= 8 \frac{d}{d t} (t^{3})

Apply the Power Rule:

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

= 8 \cdot 3 t^{3 - 1}

Simplify

= 24 t^{2}

\frac{d}{d x} (ln (x^{2} + x^{2} + 1))

\int \frac{1}{1 - 2 x ^{2}} d x

\frac{d}{d x} (x^{\frac{1}{3}} + 2 x)

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

f (x) = x^{4} + 2 x^{2}, at x = 1