Question

question
find (dy) and evaluate when (x = - 3) and (dx = 0.4) for the function (y=\frac{x^{3}}{5x + 5}).
(enter an exact answer.)
provide your answer below:
(dy=square)

Explanation:

Step1: Differentiate the function

First, rewrite $y=\frac{x^{3}}{5x + 5}$ as $y=\frac{1}{5}\cdot\frac{x^{3}}{x + 1}$. Use the quotient - rule, if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = x^{3}$, $u^\prime=3x^{2}$, $v=x + 1$, $v^\prime = 1$. So $y^\prime=\frac{1}{5}\cdot\frac{3x^{2}(x + 1)-x^{3}\cdot1}{(x + 1)^{2}}=\frac{1}{5}\cdot\frac{3x^{3}+3x^{2}-x^{3}}{(x + 1)^{2}}=\frac{1}{5}\cdot\frac{2x^{3}+3x^{2}}{(x + 1)^{2}}$.

Step2: Recall the relationship between $dy$ and $dx$

We know that $dy=y^\prime dx$. Substitute $y^\prime$ and the given values of $x$ and $dx$. When $x=-3$ and $dx = 0.4$, first find $y^\prime$ at $x=-3$.
$y^\prime=\frac{1}{5}\cdot\frac{2(-3)^{3}+3(-3)^{2}}{(-3 + 1)^{2}}=\frac{1}{5}\cdot\frac{2(-27)+3(9)}{4}=\frac{1}{5}\cdot\frac{-54 + 27}{4}=\frac{1}{5}\cdot\frac{-27}{4}=-\frac{27}{20}$.

Step3: Calculate $dy$

Since $dy=y^\prime dx$, substitute $y^\prime=-\frac{27}{20}$ and $dx = 0.4$. Then $dy=-\frac{27}{20}\times0.4=-\frac{27}{20}\times\frac{4}{10}=-\frac{27}{50}=-0.54$.

Answer:

$-0.54$