What is the derivative of P(x)=-1000+12x+0.0004x^2 ?

The derivative of P(x)=-1000+12x+0.0004x^2 is 0.0008x+12

What is the first derivative of P(x)=-1000+12x+0.0004x^2 ?

The first derivative of P(x)=-1000+12x+0.0004x^2 is 0.0008x+12

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derivative of

P (x) = - 1000 + 12 x + 0.0004 x^{2}

0.0008 x + 12

Solution steps

\frac{d}{d x} (- 1000 + 12 x + 0.0004 x^{2})

Apply the Sum/Difference Rule:

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

= - \frac{d}{d x} (1000) + \frac{d}{d x} (12 x) + \frac{d}{d x} (0.0004 x^{2})

\frac{d}{d x} (1000) = 0

\frac{d}{d x} (12 x) = 12

\frac{d}{d x} (0.0004 x^{2}) = 0.0008 x

= - 0 + 12 + 0.0008 x

Simplify

= 0.0008 x + 12

x^{3} (x - 5)^{2}

0.05 \cdot x \cdot ln (\frac{10}{x}), x = 1

\frac{5 s}{s ^{2} + 2 s + 5}

\int 3 cos^{2} (x) sin^{2} (x) d x

x \to 0 lim (x^{3 x})

What is the derivative of P(x)=-1000+12x+0.0004x^2 ?
The derivative of P(x)=-1000+12x+0.0004x^2 is 0.0008x+12
What is the first derivative of P(x)=-1000+12x+0.0004x^2 ?
The first derivative of P(x)=-1000+12x+0.0004x^2 is 0.0008x+12