What is the first d/(dt)(at^2+bt+c) ?

The first d/(dt)(at^2+bt+c) is 2at+b

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d/(dt)(at^2+bt+c)

\frac{d}{d t} (a t^{2} + b t + c)

2 a t + b

Solution steps

\frac{d}{d t} (a t^{2} + b t + c)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (a t^{2}) + \frac{d}{d t} (b t) + \frac{d c}{d t}

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

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

\frac{d c}{d t} = 0

= 2 a t + b + 0

Simplify

= 2 a t + b