What is the first d/(dt)(6-3t^2) ?

The first d/(dt)(6-3t^2) is -6t

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d/(dt)(6-3t^2)

\frac{d}{d t} (6 - 3 t^{2})

- 6 t

Solution steps

\frac{d}{d t} (6 - 3 t^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (6) - \frac{d}{d t} (3 t^{2})

\frac{d}{d t} (6) = 0

\frac{d}{d t} (3 t^{2}) = 6 t

= 0 - 6 t

Simplify

= - 6 t

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