question 18 · 1 point\nfind (dy) and evaluate when (x = - 2) and (dx = 0.1) for the function…

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = x^{4}+3$, so $u^\prime = 4x^{3}$, and $v = 3x + 8$, so $v^\prime=3$. Then $y^\prime=\frac{4x^{3}(3x + 8)-(x^{4}+3)\times3}{(3x + 8)^{2}}=\frac{12x^{4}+32x^{3}-3x^{4}-9}{(3x + 8)^{2}}=\frac{9x^{4}+32x^{3}-9}{(3x + 8)^{2}}$.

Step2: Recall the definition of $dy$

We know that $dy=y^\prime dx$. So $dy=\frac{9x^{4}+32x^{3}-9}{(3x + 8)^{2}}dx$.

Step3: Substitute $x=-2$ and $dx = 0.1$

First, substitute $x = - 2$ into the derivative: [ \begin{align*} y^\prime&=\frac{9(-2)^{4}+32(-2)^{3}-9}{(3(-2)+8)^{2}}\ &=\frac{9\times16+32\times(-8)-9}{(-6 + 8)^{2}}\ &=\frac{144-256-9}{4}\ &=\frac{144-265}{4}\ &=-\frac{121}{4} \end{align*} ] Then, since $dy=y^\prime dx$ and $dx = 0.1=\frac{1}{10}$, we have $dy=-\frac{121}{4}\times\frac{1}{10}=-\frac{121}{40}$.

Answer:

$-\frac{121}{40}$