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

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

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

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

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

3 x^{2} + 8

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 3 x^{2} + 8

\frac{1}{1 - x ^{2}}, 0

y = 2 x, y = x, x = 0, x = 4

f (x) = 5 - 2 x^{2}, (1, 3)

\frac{d}{d x} (7 e^{- x} - 8 e^{- 7 x})

f (x) = \frac{( x + 3 )}{2 x + 1}, at x = - 1

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