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

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

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

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

2 - 2 t

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 2 - 2 t

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