What is the first d/(dt)(10+5t+2t^2) ?

The first d/(dt)(10+5t+2t^2) is 4t+5

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d/(dt)(10+5t+2t^2)

\frac{d}{d t} (10 + 5 t + 2 t^{2})

4 t + 5

Solution steps

\frac{d}{d t} (10 + 5 t + 2 t^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (10) + \frac{d}{d t} (5 t) + \frac{d}{d t} (2 t^{2})

\frac{d}{d t} (10) = 0

\frac{d}{d t} (5 t) = 5

\frac{d}{d t} (2 t^{2}) = 4 t

= 0 + 5 + 4 t

Simplify

= 4 t + 5