What is the derivative of 160+3.2x-0.03x^2 ?

The derivative of 160+3.2x-0.03x^2 is -0.06x+3.2

What is the first derivative of 160+3.2x-0.03x^2 ?

The first derivative of 160+3.2x-0.03x^2 is -0.06x+3.2

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derivative of

160 + 3.2 x - 0.03 x^{2}

- 0.06 x + 3.2

Solution steps

\frac{d}{d x} (160 + 3.2 x - 0.03 x^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (160) + \frac{d}{d x} (3.2 x) - \frac{d}{d x} (0.03 x^{2})

\frac{d}{d x} (160) = 0

\frac{d}{d x} (3.2 x) = 3.2

\frac{d}{d x} (0.03 x^{2}) = 0.06 x

= 0 + 3.2 - 0.06 x

Simplify

= - 0.06 x + 3.2

\frac{d}{d x} (9 x sin (π x))

n = 0 \sum \infty (- \frac{1}{4})^{n}

f (x) = x - x^{2}, at x = - 4

\int cos (x) cos^{2} (x) d x

3 x + 4

What is the derivative of 160+3.2x-0.03x^2 ?
The derivative of 160+3.2x-0.03x^2 is -0.06x+3.2
What is the first derivative of 160+3.2x-0.03x^2 ?
The first derivative of 160+3.2x-0.03x^2 is -0.06x+3.2