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

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

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

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

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

2 x + 8

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 2 x + 8

\frac{\partial}{\partial x} (x e^{- 2 x})

(3 x - 2 x^{2})^{3}

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

(- 3.5) (- 2.2)

x \to 9 - lim (x - 9)