What is the derivative of 1/3 x^3-5/2 x^2+6x+1 ?

The derivative of 1/3 x^3-5/2 x^2+6x+1 is x^2-5x+6

What is the first derivative of 1/3 x^3-5/2 x^2+6x+1 ?

The first derivative of 1/3 x^3-5/2 x^2+6x+1 is x^2-5x+6

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

x^{2} - 5 x + 6

Solution steps

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

Simplify

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

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (\frac{5 x ^{2}}{2}) = 5 x

\frac{d}{d x} (6 x) = 6

\frac{d}{d x} (1) = 0

= x^{2} - 5 x + 6 + 0

Simplify

= x^{2} - 5 x + 6

\int - \frac{1}{x} \cdot \frac{1}{x} d x

f (x) = x - 1

\int θ sec (θ) tan (θ) d θ

(\frac{f ( x ) + x - 6}{4 - 2 x})^{'}

\int (9 - x^{2}) d x

What is the derivative of 1/3 x^3-5/2 x^2+6x+1 ?
The derivative of 1/3 x^3-5/2 x^2+6x+1 is x^2-5x+6
What is the first derivative of 1/3 x^3-5/2 x^2+6x+1 ?
The first derivative of 1/3 x^3-5/2 x^2+6x+1 is x^2-5x+6