What is the derivative of (4x-4(3x^3-x^2+1)) ?

The derivative of (4x-4(3x^3-x^2+1)) is 48x^3-48x^2+8x+4

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

The first derivative of (4x-4(3x^3-x^2+1)) is 48x^3-48x^2+8x+4

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

48 x^{3} - 48 x^{2} + 8 x + 4

Solution steps

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

Apply the Product Rule:

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

f = 4 x - 4, g = 3 x^{3} - x^{2} + 1

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

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

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

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

Simplify

4 (3 x^{3} - x^{2} + 1) + (9 x^{2} - 2 x) (4 x - 4) : 48 x^{3} - 48 x^{2} + 8 x + 4

= 48 x^{3} - 48 x^{2} + 8 x + 4

y = \frac{1}{16 x ^{2}}, y = x, y = \frac{x}{18}

f (x) = sin (cos (4 x))

\int 4 x cos (4 x) d x

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

f (x) = 1.5 e^{0.76 x}

What is the derivative of (4x-4(3x^3-x^2+1)) ?
The derivative of (4x-4(3x^3-x^2+1)) is 48x^3-48x^2+8x+4
What is the first derivative of (4x-4(3x^3-x^2+1)) ?
The first derivative of (4x-4(3x^3-x^2+1)) is 48x^3-48x^2+8x+4