What is the d/(dt)(12+8t-4t^2) ?

The d/(dt)(12+8t-4t^2) is -8t+8

What is the first d/(dt)(12+8t-4t^2) ?

The first d/(dt)(12+8t-4t^2) is -8t+8

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\frac{d}{d t} (12 + 8 t - 4 t^{2})

- 8 t + 8

Solution steps

\frac{d}{d t} (12 + 8 t - 4 t^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (12) + \frac{d}{d t} (8 t) - \frac{d}{d t} (4 t^{2})

\frac{d}{d t} (12) = 0

\frac{d}{d t} (8 t) = 8

\frac{d}{d t} (4 t^{2}) = 8 t

= 0 + 8 - 8 t

Simplify

= - 8 t + 8

\frac{d}{d x} (+ 5 - 5 x^{- 1} + x^{2} + 3)

f (x) = - 2 x - 4

\frac{- 1}{( x - 2 ) ^{2}}

5 (t) i + (40 - 5 (t)^{2}) j

2 + cos (x)