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

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

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

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

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

- 0.2 x + 40

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= - 0.2 x + 40

(2 x + 1)^{5}

y = x^{\frac{2}{5}} + 6 x

\int 6 e^{t} d t

\frac{d}{d x} (\frac{2 - 2 x - x ^{2}}{x ^{2} - 4})

\frac{1}{2} x^{2} ln^{2} (x) - 2 x^{2} ln (x) + \frac{5}{2} x^{2}

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