What is the derivative of-16x^2+16x+320 ?

The derivative of-16x^2+16x+320 is -32x+16

What is the first derivative of-16x^2+16x+320 ?

The first derivative of-16x^2+16x+320 is -32x+16

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\frac{d}{d x} (- 16 x^{2} + 16 x + 320)

- 32 x + 16

Solution steps

\frac{d}{d x} (- 16 x^{2} + 16 x + 320)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d x} (16 x^{2}) + \frac{d}{d x} (16 x) + \frac{d}{d x} (320)

\frac{d}{d x} (16 x^{2}) = 32 x

\frac{d}{d x} (16 x) = 16

\frac{d}{d x} (320) = 0

= - 32 x + 16 + 0

Simplify

= - 32 x + 16

\frac{d}{d x} (- 25 cos (5 x))

\frac{d}{d x} (2 x e^{5 x})

8 x^{9}

\int_{- 1}^{1} (1 - ∣ x ∣) d x

\frac{d}{d x} (3 x + 4 x^{2})

What is the derivative of-16x^2+16x+320 ?
The derivative of-16x^2+16x+320 is -32x+16
What is the first derivative of-16x^2+16x+320 ?
The first derivative of-16x^2+16x+320 is -32x+16