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

The derivative of f(x)=(x^3+2x^2-4)/3 is x^2+(4x)/3

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

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

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derivative of

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

x^{2} + \frac{4 x}{3}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{3} \frac{d}{d x} (x^{3} + 2 x^{2} - 4)

Apply the Sum/Difference Rule:

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

= \frac{1}{3} (\frac{d}{d x} (x^{3}) + \frac{d}{d x} (2 x^{2}) - \frac{d}{d x} (4))

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

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

\frac{d}{d x} (4) = 0

= \frac{1}{3} (3 x^{2} + 4 x - 0)

Simplify

\frac{1}{3} (3 x^{2} + 4 x - 0) : x^{2} + \frac{4 x}{3}

= x^{2} + \frac{4 x}{3}

\int (x^{2} + 1) e^{x^{3} + 3 x} d x

\int_{0}^{1} \frac{1}{x ^{0.4}} d x

\frac{1}{2} (8 t + 3)^{- \frac{1}{2}}

\frac{d}{d x} (f (x) \cdot f (x) \cdot f (x))

y^{^{'}} - \frac{8}{x} y = y^{3} \frac{1}{x ^{12}}

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