Question

find y’ by (a) applying the product rule and (b) multiplying the factors to produce a sum of simpler terms to differentiate. y=(7 - x²)(x³ - 4x + 1)
a. apply the product rule. let u=(7 - x²) and v=(x³ - 4x + 1).
\frac{d}{dx}(uv)=(7 - x²)(\square)+(x³ - 4x + 1)(\square)

Explanation:

Step1: Recall the Product Rule

The Product Rule states that if $y = uv$, then $y'=\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$. Here, $u = 7 - x^{2}$ and $v=x^{3}-4x + 1$.

Step2: Differentiate $u$ and $v$

Differentiate $u$ with respect to $x$: $\frac{du}{dx}=\frac{d}{dx}(7 - x^{2})=-2x$. Differentiate $v$ with respect to $x$: $\frac{dv}{dx}=\frac{d}{dx}(x^{3}-4x + 1)=3x^{2}-4$.

Step3: Apply the Product Rule

Substitute $u$, $v$, $\frac{du}{dx}$ and $\frac{dv}{dx}$ into the Product - Rule formula:
$\frac{d}{dx}(uv)=(7 - x^{2})(3x^{2}-4)+(x^{3}-4x + 1)(-2x)$.
Expand the expression:
\[

$$\begin{align*} &(7 - x^{2})(3x^{2}-4)+(x^{3}-4x + 1)(-2x)\\ =&7\times(3x^{2}-4)-x^{2}(3x^{2}-4)-2x\times x^{3}+8x^{2}-2x\\ =&21x^{2}-28 - 3x^{4}+4x^{2}-2x^{4}+8x^{2}-2x\\ =&- 5x^{4}+33x^{2}-2x - 28 \end{align*}$$

\]

Step4: Multiply the factors first

Multiply $y=(7 - x^{2})(x^{3}-4x + 1)$:
\[

$$\begin{align*} y&=7(x^{3}-4x + 1)-x^{2}(x^{3}-4x + 1)\\ &=7x^{3}-28x + 7-x^{5}+4x^{3}-x^{2}\\ &=-x^{5}+11x^{3}-x^{2}-28x + 7 \end{align*}$$

\]
Differentiate $y$ with respect to $x$: $y'=-5x^{4}+33x^{2}-2x - 28$.

Answer:

$y'=-5x^{4}+33x^{2}-2x - 28$