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

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

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

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

3 t^{2} + 3

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 3 t^{2} + 3