What is the first d/(dt)(t^2+3t-2) ?

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

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d/(dt)(t^2+3t-2)

\frac{d}{d t} (t^{2} + 3 t - 2)

2 t + 3

Solution steps

\frac{d}{d t} (t^{2} + 3 t - 2)

Apply the Sum/Difference Rule:

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

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

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

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

\frac{d}{d t} (2) = 0

= 2 t + 3 - 0

Simplify

= 2 t + 3

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