What is the derivative of f(x)=x^2+8x ?

The derivative of f(x)=x^2+8x is 2x+8

What is the first derivative of f(x)=x^2+8x ?

The first derivative of f(x)=x^2+8x is 2x+8

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derivative of

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

x \to \infty lim (x^{- x})

\int_{0}^{2} x (x^{2} + 1)^{5} d x

v^{2} (2 v + 1)

\frac{d}{d x} (\frac{x ^{2} ln ( x )}{e ^{x}})

What is the derivative of f(x)=x^2+8x ?
The derivative of f(x)=x^2+8x is 2x+8
What is the first derivative of f(x)=x^2+8x ?
The first derivative of f(x)=x^2+8x is 2x+8