What is the derivative of 2/3 x^3-4x^2+10x+6 ?

The derivative of 2/3 x^3-4x^2+10x+6 is 2x^2-8x+10

What is the first derivative of 2/3 x^3-4x^2+10x+6 ?

The first derivative of 2/3 x^3-4x^2+10x+6 is 2x^2-8x+10

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\frac{d}{d x} (\frac{2}{3} x^{3} - 4 x^{2} + 10 x + 6)

2 x^{2} - 8 x + 10

Solution steps

\frac{d}{d x} (\frac{2}{3} x^{3} - 4 x^{2} + 10 x + 6)

Simplify

\frac{2}{3} x^{3} - 4 x^{2} + 10 x + 6 : \frac{2 x ^{3}}{3} - 4 x^{2} + 10 x + 6

= \frac{d}{d x} (\frac{2 x ^{3}}{3} - 4 x^{2} + 10 x + 6)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{2 x ^{3}}{3}) - \frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (10 x) + \frac{d}{d x} (6)

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

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

\frac{d}{d x} (10 x) = 10

\frac{d}{d x} (6) = 0

= 2 x^{2} - 8 x + 10 + 0

Simplify

= 2 x^{2} - 8 x + 10

\frac{d y}{d x} = \frac{y}{x}, y (1) = 2

\int_{0}^{10} 200 - 4 x d x

y^{^{'}} = 8 x y, y (1) = 1

in v erse l a pl a ce \frac{1}{( s + \frac{2}{7} ) ^{2}} \cdot \frac{1}{s}

\frac{\partial}{\partial u} (\frac{u + v}{2})

What is the derivative of 2/3 x^3-4x^2+10x+6 ?
The derivative of 2/3 x^3-4x^2+10x+6 is 2x^2-8x+10
What is the first derivative of 2/3 x^3-4x^2+10x+6 ?
The first derivative of 2/3 x^3-4x^2+10x+6 is 2x^2-8x+10