What is the first d/(dt)(1/6 t^3) ?

The first d/(dt)(1/6 t^3) is (t^2)/2

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d/(dt)(1/6 t^3)

\frac{d}{d t} (\frac{1}{6} t^{3})

\frac{t ^{2}}{2}

Solution steps

\frac{d}{d t} (\frac{1}{6} t^{3})

Take the constant out:

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

= \frac{1}{6} \frac{d}{d t} (t^{3})

Apply the Power Rule:

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

= \frac{1}{6} \cdot 3 t^{3 - 1}

Simplify

\frac{1}{6} \cdot 3 t^{3 - 1} : \frac{t ^{2}}{2}

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

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