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

The derivative of x^2+x+8 is 2x+1

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

The first derivative of x^2+x+8 is 2x+1

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derivative of

x^{2} + x + 8

2 x + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d x}{d x} = 1

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

= 2 x + 1 + 0

Simplify

= 2 x + 1

f (x) = ln (e^{x})

\int (e^{- 2 x}) d x

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

\int \frac{1}{x ^{2} + 6 x + 13} d x

\int (\frac{1}{x}) d x

What is the derivative of x^2+x+8 ?
The derivative of x^2+x+8 is 2x+1
What is the first derivative of x^2+x+8 ?
The first derivative of x^2+x+8 is 2x+1