What is the derivative of f(x)=-12x^2+400x-3800 ?

The derivative of f(x)=-12x^2+400x-3800 is -24x+400

What is the first derivative of f(x)=-12x^2+400x-3800 ?

The first derivative of f(x)=-12x^2+400x-3800 is -24x+400

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derivative of

f (x) = - 12 x^{2} + 400 x - 3800

- 24 x + 400

Solution steps

\frac{d}{d x} (- 12 x^{2} + 400 x - 3800)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d x} (12 x^{2}) + \frac{d}{d x} (400 x) - \frac{d}{d x} (3800)

\frac{d}{d x} (12 x^{2}) = 24 x

\frac{d}{d x} (400 x) = 400

\frac{d}{d x} (3800) = 0

= - 24 x + 400 - 0

Simplify

= - 24 x + 400

3 x - 4 x^{2}

\int (1 - sin^{2} (x)) cos (x) d x

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

\frac{\partial}{\partial x} (2 x^{2} y + sin (x^{4}))

x \to 1 lim (x^{\frac{8}{1 - x}})

What is the derivative of f(x)=-12x^2+400x-3800 ?
The derivative of f(x)=-12x^2+400x-3800 is -24x+400
What is the first derivative of f(x)=-12x^2+400x-3800 ?
The first derivative of f(x)=-12x^2+400x-3800 is -24x+400