What is the derivative of f(x)=(x^3)/3+4x^2+12x+19 ?

The derivative of f(x)=(x^3)/3+4x^2+12x+19 is x^2+8x+12

What is the first derivative of f(x)=(x^3)/3+4x^2+12x+19 ?

The first derivative of f(x)=(x^3)/3+4x^2+12x+19 is x^2+8x+12

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derivative of

f (x) = \frac{x ^{3}}{3} + 4 x^{2} + 12 x + 19

x^{2} + 8 x + 12

Solution steps

\frac{d}{d x} (\frac{x ^{3}}{3} + 4 x^{2} + 12 x + 19)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{3}}{3}) + \frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (12 x) + \frac{d}{d x} (19)

\frac{d}{d x} (\frac{x ^{3}}{3}) = x^{2}

\frac{d}{d x} (4 x^{2}) = 8 x

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

\frac{d}{d x} (19) = 0

= x^{2} + 8 x + 12 + 0

Simplify

= x^{2} + 8 x + 12

\int - x cos (x) d x

\frac{\partial}{\partial x} (A e^{- 3 x})

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 1, y^{^{'}} (0) = - 5

f (x) = arccos (e^{2 x})

x^{3} - 8 x^{2} + 12 x, - x^{3} + 8 x^{2} - 12 x

What is the derivative of f(x)=(x^3)/3+4x^2+12x+19 ?
The derivative of f(x)=(x^3)/3+4x^2+12x+19 is x^2+8x+12
What is the first derivative of f(x)=(x^3)/3+4x^2+12x+19 ?
The first derivative of f(x)=(x^3)/3+4x^2+12x+19 is x^2+8x+12