What is the derivative of x^2+4x-8 ?

The derivative of x^2+4x-8 is 2x+4

What is the first derivative of x^2+4x-8 ?

The first derivative of x^2+4x-8 is 2x+4

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\frac{d}{d x} (x^{2} + 4 x - 8)

2 x + 4

Solution steps

\frac{d}{d x} (x^{2} + 4 x - 8)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{2}) + \frac{d}{d x} (4 x) - \frac{d}{d x} (8)

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

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

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

= 2 x + 4 - 0

Simplify

= 2 x + 4

x \to 0 lim (\frac{( e ^{3 x} - 1 )}{sin ( 5 x )})

\int cos^{3} (x) sin^{3} (x) d x

y^{^{''}} - 81 y = 0

e^{2 x + 4}

\frac{2 x + 3}{3 x - 2}

What is the derivative of x^2+4x-8 ?
The derivative of x^2+4x-8 is 2x+4
What is the first derivative of x^2+4x-8 ?
The first derivative of x^2+4x-8 is 2x+4