What is the derivative of y=(x^3+3x^2-5x+2)/3 ?

The derivative of y=(x^3+3x^2-5x+2)/3 is x^2+2x-5/3

What is the first derivative of y=(x^3+3x^2-5x+2)/3 ?

The first derivative of y=(x^3+3x^2-5x+2)/3 is x^2+2x-5/3

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derivative of

y = \frac{x ^{3} + 3 x ^{2} - 5 x + 2}{3}

x^{2} + 2 x - \frac{5}{3}

Solution steps

\frac{d}{d x} (\frac{x ^{3} + 3 x ^{2} - 5 x + 2}{3})

Take the constant out:

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

= \frac{1}{3} \frac{d}{d x} (x^{3} + 3 x^{2} - 5 x + 2)

Apply the Sum/Difference Rule:

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

= \frac{1}{3} (\frac{d}{d x} (x^{3}) + \frac{d}{d x} (3 x^{2}) - \frac{d}{d x} (5 x) + \frac{d}{d x} (2))

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

\frac{d}{d x} (3 x^{2}) = 6 x

\frac{d}{d x} (5 x) = 5

\frac{d}{d x} (2) = 0

= \frac{1}{3} (3 x^{2} + 6 x - 5 + 0)

Simplify

\frac{1}{3} (3 x^{2} + 6 x - 5 + 0) : x^{2} + 2 x - \frac{5}{3}

= x^{2} + 2 x - \frac{5}{3}

y^{^{'}} - 8 y^{3} = 0

\int \frac{1}{2 x - 1 - 4 2 x - 1} d x

\int x^{1.1} (\frac{1}{3 x} - 1) d x

x^{2} + y^{2} = 25

\int_{1}^{2} x (x - 1)^{9} d x

What is the derivative of y=(x^3+3x^2-5x+2)/3 ?
The derivative of y=(x^3+3x^2-5x+2)/3 is x^2+2x-5/3
What is the first derivative of y=(x^3+3x^2-5x+2)/3 ?
The first derivative of y=(x^3+3x^2-5x+2)/3 is x^2+2x-5/3