What is the derivative of y=-0.0959x^2+106.83x+71149 ?

The derivative of y=-0.0959x^2+106.83x+71149 is -0.1918x+106.83

What is the first derivative of y=-0.0959x^2+106.83x+71149 ?

The first derivative of y=-0.0959x^2+106.83x+71149 is -0.1918x+106.83

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derivative of

y = - 0.0959 x^{2} + 106.83 x + 71149

- 0.1918 x + 106.83

Solution steps

\frac{d}{d x} (- 0.0959 x^{2} + 106.83 x + 71149)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d x} (0.0959 x^{2}) + \frac{d}{d x} (106.83 x) + \frac{d}{d x} (71149)

\frac{d}{d x} (0.0959 x^{2}) = 0.1918 x

\frac{d}{d x} (106.83 x) = 106.83

\frac{d}{d x} (71149) = 0

= - 0.1918 x + 106.83 + 0

Simplify

= - 0.1918 x + 106.83

\int 2 e^{t x} x d x

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\int_{0}^{1} 2 π x (5 x (x - 1)^{2}) d x

\frac{5 - sec ( x )}{tan ( x )}

What is the derivative of y=-0.0959x^2+106.83x+71149 ?
The derivative of y=-0.0959x^2+106.83x+71149 is -0.1918x+106.83
What is the first derivative of y=-0.0959x^2+106.83x+71149 ?
The first derivative of y=-0.0959x^2+106.83x+71149 is -0.1918x+106.83