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

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

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

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

3 t^{2}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d t} (t^{3}) = 3 t^{2}

\frac{d}{d t} (1) = 0

= 3 t^{2} + 0

Simplify

= 3 t^{2}

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