What is the derivative of x^2+x+4 ?

The derivative of x^2+x+4 is 2x+1

What is the first derivative of x^2+x+4 ?

The first derivative of x^2+x+4 is 2x+1

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

2 x + 1

Solution steps

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

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} (4)

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

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

\frac{d}{d x} (4) = 0

= 2 x + 1 + 0

Simplify

= 2 x + 1

\frac{d h}{d t} = 4 - 16 t^{2}, h (1) = 0

\frac{d}{d x} (ln (x^{2} - 16 x))

\int \frac{x ^{2} + 3 x - 4}{x ^{2} - 2 x - 8} d x

\int (y^{2} - 1) d y

x^{3} - 13 x^{2} + 40 x, - x^{3} + 13 x^{2} - 40 x

What is the derivative of x^2+x+4 ?
The derivative of x^2+x+4 is 2x+1
What is the first derivative of x^2+x+4 ?
The first derivative of x^2+x+4 is 2x+1