What is the derivative of-8x^2+500x+140 ?

The derivative of-8x^2+500x+140 is -16x+500

What is the first derivative of-8x^2+500x+140 ?

The first derivative of-8x^2+500x+140 is -16x+500

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derivative of

- 8 x^{2} + 500 x + 140

- 16 x + 500

Solution steps

\frac{d}{d x} (- 8 x^{2} + 500 x + 140)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d x} (8 x^{2}) + \frac{d}{d x} (500 x) + \frac{d}{d x} (140)

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

\frac{d}{d x} (500 x) = 500

\frac{d}{d x} (140) = 0

= - 16 x + 500 + 0

Simplify

= - 16 x + 500

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

\int (2 - x) 4 - x^{2} d x

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

\int_{0}^{2} 9 x^{3} x^{2} + 4 d x

2 x^{2} - x - 3

What is the derivative of-8x^2+500x+140 ?
The derivative of-8x^2+500x+140 is -16x+500
What is the first derivative of-8x^2+500x+140 ?
The first derivative of-8x^2+500x+140 is -16x+500