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

The derivative of g(x)=-8x^2-x+8 is -16x-1

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

The first derivative of g(x)=-8x^2-x+8 is -16x-1

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derivative of

g (x) = - 8 x^{2} - x + 8

- 16 x - 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (8 x^{2}) = 16 x

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

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

= - 16 x - 1 + 0

Simplify

= - 16 x - 1

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

ln (t^{2})

\frac{ln ( 4 x )}{4 x}

x \to 0 lim (\frac{e ^{- 2 x} - 1}{x})

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

What is the derivative of g(x)=-8x^2-x+8 ?
The derivative of g(x)=-8x^2-x+8 is -16x-1
What is the first derivative of g(x)=-8x^2-x+8 ?
The first derivative of g(x)=-8x^2-x+8 is -16x-1