Questions: Differentiate. Find f'(x) for f(x)=(2x-4)(2x^3-x^2+1). (Use the Product Rule.) f'(x)=16x^3-10x^2+30x+2 f'(x)=12x^3+30x^2-10x+2 f'(x)=16x^3-30x^2+8x+2 f'(x)=4x^3-10x^2-30x+2

Transcript text: Differentiate. Find $f^{\prime}(x)$ for $f(x)=(2 x-4)\left(2 x^{3}-x^{2}+1\right)$. (Use the Product Rule.) $f^{\prime}(x)=16 x^{3}-10 x^{2}+30 x+2$ $f^{\prime}(x)=12 x^{3}+30 x^{2}-10 x+2$ $f^{\prime}(x)=16 x^{3}-30 x^{2}+8 x+2$ $f^{\prime}(x)=4 x^{3}-10 x^{2}-30 x+2$

Solution

Solution Steps

To differentiate the function $ f(x) = (2x - 4)(2x^3 - x^2 + 1) $, we will use the Product Rule. The Product Rule states that if you have a function $ f(x) = u(x) \cdot v(x) $, then the derivative $ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) $. Here, $ u(x) = 2x - 4 $ and $ v(x) = 2x^3 - x^2 + 1 $. We will find the derivatives $ u'(x) $ and $ v'(x) $, and then apply the Product Rule.

Step 1: Identify Functions

We start with the function $ f(x) = (2x - 4)(2x^3 - x^2 + 1) $. Here, we define:

$ u(x) = 2x - 4 $
$ v(x) = 2x^3 - x^2 + 1 $

Step 2: Compute Derivatives

Next, we compute the derivatives of $ u $ and $ v $:

$ u'(x) = 2 $
$ v'(x) = 6x^2 - 2x $

Step 3: Apply the Product Rule

Using the Product Rule, we find the derivative $ f'(x) $: \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \] Substituting the values we have: \[ f'(x) = 2(2x^3 - x^2 + 1) + (2x - 4)(6x^2 - 2x) \]

Step 4: Simplify the Expression

Now we simplify the expression: \[ f'(x) = 4x^3 - 2x^2 + (2x - 4)(6x^2 - 2x) \] Expanding $ (2x - 4)(6x^2 - 2x) $: \[ = 12x^3 - 4x^2 - 24x^2 + 8x = 12x^3 - 28x^2 + 8x \] Combining all terms: \[ f'(x) = 4x^3 - 2x^2 + 12x^3 - 28x^2 + 8x = 16x^3 - 30x^2 + 8x + 2 \]