differentiate the function. y=(x + 8)(x³ + 5x + 1) y=□

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u=x + 8$ and $v=x^{3}+5x + 1$.

Step2: Differentiate $u$ and $v$

Differentiate $u$ with respect to $x$: $u'=\frac{d}{dx}(x + 8)=1$. Differentiate $v$ with respect to $x$: $v'=\frac{d}{dx}(x^{3}+5x + 1)=3x^{2}+5$.

Step3: Substitute into product - rule

$y'=u'v+uv'=1\cdot(x^{3}+5x + 1)+(x + 8)\cdot(3x^{2}+5)$.

Step4: Expand the expression

[ \begin{align*} y'&=x^{3}+5x + 1+(x\cdot3x^{2}+x\cdot5+8\cdot3x^{2}+8\cdot5)\ &=x^{3}+5x + 1+(3x^{3}+5x+24x^{2}+40)\ &=x^{3}+5x + 1+3x^{3}+5x+24x^{2}+40\ &=4x^{3}+24x^{2}+10x + 41 \end{align*} ]

Answer:

$4x^{3}+24x^{2}+10x + 41$