What is the derivative of 6t^2+8 ?

The derivative of 6t^2+8 is 12t

What is the first derivative of 6t^2+8 ?

The first derivative of 6t^2+8 is 12t

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derivative of

6 t^{2} + 8

12 t

Solution steps

\frac{d}{d t} (6 t^{2} + 8)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (6 t^{2}) + \frac{d}{d t} (8)

\frac{d}{d t} (6 t^{2}) = 12 t

\frac{d}{d t} (8) = 0

= 12 t + 0

Simplify

= 12 t

\int \frac{14}{14 + e ^{x}} d x

f (t) = t^{3} - 2 t^{2} + 5 t + 4

f (x) = \frac{e ^{- x}}{x + 1}, at x = 1

\frac{d}{d x} (\frac{x}{( 1 - x )})

\int x^{3} x - 4 d x