What is the derivative of-0.1x^2-x+40 ?

The derivative of-0.1x^2-x+40 is -0.2x-1

What is the first derivative of-0.1x^2-x+40 ?

The first derivative of-0.1x^2-x+40 is -0.2x-1

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derivative of

- 0.1 x^{2} - x + 40

- 0.2 x - 1

Solution steps

\frac{d}{d x} (- 0.1 x^{2} - x + 40)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d x} (0.1 x^{2}) - \frac{d x}{d x} + \frac{d}{d x} (40)

\frac{d}{d x} (0.1 x^{2}) = 0.2 x

\frac{d x}{d x} = 1

\frac{d}{d x} (40) = 0

= - 0.2 x - 1 + 0

Simplify

= - 0.2 x - 1

\int 7 x e^{(3 x - 1)} d x

n = 1 \sum \infty \frac{n !}{10 8 ^{n}}

y + x y^{^{'}} = a (1 + x y)

\frac{d}{d x} ((2 x + 5)^{7})

y^{^{'}} = \frac{12 ( 2 y - 1 )}{( 3 x - 2 ) ^{2}}, y (2) = 0

What is the derivative of-0.1x^2-x+40 ?
The derivative of-0.1x^2-x+40 is -0.2x-1
What is the first derivative of-0.1x^2-x+40 ?
The first derivative of-0.1x^2-x+40 is -0.2x-1