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

The derivative of f(x)=(2x-4)(2x^3-x^2+1) is 16x^3-30x^2+8x+2

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

The first derivative of f(x)=(2x-4)(2x^3-x^2+1) is 16x^3-30x^2+8x+2

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derivative of

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

16 x^{3} - 30 x^{2} + 8 x + 2

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{d}{d x} (2 x - 4) = 2

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

= 2 (2 x^{3} - x^{2} + 1) + (6 x^{2} - 2 x) (2 x - 4)

Simplify

2 (2 x^{3} - x^{2} + 1) + (6 x^{2} - 2 x) (2 x - 4) : 16 x^{3} - 30 x^{2} + 8 x + 2

= 16 x^{3} - 30 x^{2} + 8 x + 2

y = \frac{1}{( 1 + x ^{2} )}

h (y, w) \to 0 lim ((y - h (y, w))^{w (y)})

x \to 0 lim (\frac{( x ^{2} )}{∣ x ∣})

\frac{d y}{d x} - y = 8

(0, 2), (5, - 2)

What is the derivative of f(x)=(2x-4)(2x^3-x^2+1) ?
The derivative of f(x)=(2x-4)(2x^3-x^2+1) is 16x^3-30x^2+8x+2
What is the first derivative of f(x)=(2x-4)(2x^3-x^2+1) ?
The first derivative of f(x)=(2x-4)(2x^3-x^2+1) is 16x^3-30x^2+8x+2