What is the first derivative of 4x^2+8 ?

The first derivative of 4x^2+8 is 8x

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derivative of 4x^2+8

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

8 x

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 8 x + 0

Simplify

= 8 x

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