Question

find y’ by (a) applying the product rule and (b) multiplying the factors to produce a

y = (4 - x²)(x³ - 4x + 2)

a. apply the product rule. let u = (4 - x²) and v = (x³ - 4x + 2).

\\(\frac{d}{dx}(uv)=(4 - x²)(3x² - 4)+(x³ - 4x + 2)(- 2x)\\)

b. multiply the factors of the original expression, u and v, to produce a sum of simpler

y =
(simplify your answer.)

Explanation:

Step1: Multiply the factors

\[

$$\begin{align*} y&=(4 - x^{2})(x^{3}-4x + 2)\\ &=4(x^{3}-4x + 2)-x^{2}(x^{3}-4x + 2)\\ &=4x^{3}-16x + 8-(x^{5}-4x^{3}+2x^{2})\\ &=4x^{3}-16x + 8 - x^{5}+4x^{3}-2x^{2}\\ &=-x^{5}+8x^{3}-2x^{2}-16x + 8 \end{align*}$$

\]

Step2: Differentiate the result

The power - rule for differentiation is $\frac{d}{dx}(x^{n})=nx^{n - 1}$.
\[

$$\begin{align*} y'&=\frac{d}{dx}(-x^{5}+8x^{3}-2x^{2}-16x + 8)\\ &=-5x^{4}+8\times3x^{2}-2\times2x-16+0\\ &=-5x^{4}+24x^{2}-4x - 16 \end{align*}$$

\]

Answer:

$-x^{5}+8x^{3}-2x^{2}-16x + 8$ (for the expanded form of $y$) and $y'=-5x^{4}+24x^{2}-4x - 16$