What is the derivative of f(x)=((2x-2)(3x^3-x^2+1)) ?

The derivative of f(x)=((2x-2)(3x^3-x^2+1)) is 24x^3-24x^2+4x+2

What is the first derivative of f(x)=((2x-2)(3x^3-x^2+1)) ?

The first derivative of f(x)=((2x-2)(3x^3-x^2+1)) is 24x^3-24x^2+4x+2

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derivative of

f (x) = ((2 x - 2) (3 x^{3} - x^{2} + 1))

24 x^{3} - 24 x^{2} + 4 x + 2

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 2 x - 2, g = 3 x^{3} - x^{2} + 1

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

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

\frac{d}{d x} (3 x^{3} - x^{2} + 1) = 9 x^{2} - 2 x

= 2 (3 x^{3} - x^{2} + 1) + (9 x^{2} - 2 x) (2 x - 2)

Simplify

2 (3 x^{3} - x^{2} + 1) + (9 x^{2} - 2 x) (2 x - 2) : 24 x^{3} - 24 x^{2} + 4 x + 2

= 24 x^{3} - 24 x^{2} + 4 x + 2

\int_{2}^{5} ∣ x - 3 ∣ d x

(x^{2} + 1) y^{^{'}} + 8 x (y - 1) = 0

x^{4} - 8 x^{3}

\frac{\partial}{\partial x} (\frac{3}{3 x + 2 y})

\frac{d y}{d x} = \frac{x y + x ^{2} + y ^{2}}{x ^{2}}

What is the derivative of f(x)=((2x-2)(3x^3-x^2+1)) ?
The derivative of f(x)=((2x-2)(3x^3-x^2+1)) is 24x^3-24x^2+4x+2
What is the first derivative of f(x)=((2x-2)(3x^3-x^2+1)) ?
The first derivative of f(x)=((2x-2)(3x^3-x^2+1)) is 24x^3-24x^2+4x+2