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

The d/(dt)(4+12t+8t^2) is 16t+12

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

The first d/(dt)(4+12t+8t^2) is 16t+12

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

16 t + 12

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d t} (4) = 0

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

\frac{d}{d t} (8 t^{2}) = 16 t

= 0 + 12 + 16 t

Simplify

= 16 t + 12

\frac{ln ( x + 1 )}{x}

\frac{d}{d x} ((\frac{x}{( y + z )})^{2})

x^{3} + 3 x^{2} + 5

f (x) = 2 cos (2 x - 1)

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