Question

b. 2.5 pts if $\frac{dx}{dt}=3$, find $\frac{dy}{dt}$ when $x = - 2$ and $y=\frac{2}{3}sqrt{5}$. 3. (exam 2 review #9) 5 pts define $h(x)=f(x)g(x)-\frac{10}{g(x)}$. now, suppose $f(2)=7$, $f(2)= - 5$, $g(2)=3$, $g(2)= - 4$. find $h(2).

Explanation:

Step1: Recall the product - rule and quotient - rule

The product - rule states that if $u(x)=f(x)g(x)$, then $u^\prime(x)=f^\prime(x)g(x)+f(x)g^\prime(x)$. The quotient - rule states that if $v(x)=\frac{u(x)}{w(x)}$, then $v^\prime(x)=\frac{u^\prime(x)w(x)-u(x)w^\prime(x)}{w(x)^2}$. For $h(x)=f(x)g(x)-\frac{10}{g(x)}$, we can find $h^\prime(x)$ as the sum of the derivative of $f(x)g(x)$ and the derivative of $-\frac{10}{g(x)}$.

Step2: Differentiate $h(x)$

The derivative of $y = f(x)g(x)$ using the product - rule is $y^\prime=f^\prime(x)g(x)+f(x)g^\prime(x)$. The derivative of $y=-\frac{10}{g(x)}$ using the quotient - rule with $u=- 10$ (so $u^\prime = 0$) and $w = g(x)$ is $y^\prime=\frac{0\times g(x)-(-10)\times g^\prime(x)}{(g(x))^2}=\frac{10g^\prime(x)}{(g(x))^2}$. So, $h^\prime(x)=f^\prime(x)g(x)+f(x)g^\prime(x)+\frac{10g^\prime(x)}{(g(x))^2}$.

Step3: Substitute $x = 2$

Substitute $x = 2$ into $h^\prime(x)$:
\[

$$\begin{align*} h^\prime(2)&=f^\prime(2)g(2)+f(2)g^\prime(2)+\frac{10g^\prime(2)}{(g(2))^2}\\ &=(-5)\times3 + 7\times(-4)+\frac{10\times(-4)}{3^2}\\ &=-15-28-\frac{40}{9}\\ &=-43-\frac{40}{9}\\ &=\frac{-387 - 40}{9}\\ &=\frac{-427}{9} \end{align*}$$

\]

Answer:

$-\frac{427}{9}$