What is the first derivative of 3x^2+8 ?

The first derivative of 3x^2+8 is 6x

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derivative of 3x^2+8

\frac{d}{d x} (3 x^{2} + 8)

6 x

Solution steps

\frac{d}{d x} (3 x^{2} + 8)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (8)

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

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

= 6 x + 0

Simplify

= 6 x

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

x \to \frac{1}{3} x lim (sin (x))

y^{^{'}} = e^{2 x - y}

2 x^{2} + 2 x

f (x) = \frac{x}{x ^{2} + 81}, at x = 0