What is the derivative of x^3+8 ?

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

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

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

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derivative of

x^{3} + 8

3 x^{2}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 3 x^{2} + 0

Simplify

= 3 x^{2}

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

t \to 0 lim (\frac{2}{1 + t})

\frac{\partial}{\partial x} (e^{x} y)

\int_{0}^{8} (1 + 3 x) d x

\int e^{\frac{3}{x}} d x