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

\int_{0}^{π} cos (x) - sin (x) d x

\frac{\partial}{\partial x} ((x + y)^{\frac{1}{2}})

\frac{\partial}{\partial x} (4 x^{3} + y^{2})

\int x^{2} (5^{3} + 7^{2} - 6 x - 3) d x

x = - 2, x = 1, y = e^{x}, y = 3 - e^{x}