What is the derivative of-x^2+1.8x+2.5 ?

The derivative of-x^2+1.8x+2.5 is -2x+1.8

What is the first derivative of-x^2+1.8x+2.5 ?

The first derivative of-x^2+1.8x+2.5 is -2x+1.8

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\frac{d}{d x} (- x^{2} + 1.8 x + 2.5)

- 2 x + 1.8

Solution steps

\frac{d}{d x} (- x^{2} + 1.8 x + 2.5)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d x} (x^{2}) + \frac{d}{d x} (1.8 x) + \frac{d}{d x} (2.5)

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

\frac{d}{d x} (1.8 x) = 1.8

\frac{d}{d x} (2.5) = 0

= - 2 x + 1.8 + 0

Simplify

= - 2 x + 1.8

y^{^{''}} - y = 0, y (0) = 1, y^{^{'}} (0) = 1

x \to 0 - lim (ln ∣ e^{x} - 1 ∣)

\frac{d}{d x} (3 x + x^{\frac{1}{2}})

y = (x - 4) (x^{2} + 8), (1, - 27)

\frac{d}{d x} (e^{x} - 1 + e^{x} x)

What is the derivative of-x^2+1.8x+2.5 ?
The derivative of-x^2+1.8x+2.5 is -2x+1.8
What is the first derivative of-x^2+1.8x+2.5 ?
The first derivative of-x^2+1.8x+2.5 is -2x+1.8