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

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

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

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

t^{2}

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule:

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

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

Simplify

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

= t^{2}

y^{^{''}} + 25 y = t e^{6 t}

\frac{d y}{d x}, x y = 3

\frac{d}{d x} (e^{x + c})

\int \frac{1}{3 4 x + 7} d x

2 y^{^{''}} + 13 y^{^{'}} - 7 y = 0