Question

section 2.5: product and quotient rules
score: 50/80 answered: 5/8
question 6
find the following using the table below.

x	1	2	3	4
f(x)	3	2	1	4
g(x)	2	1	3	4
g(x)	1	4	3	2

h(1) if h(x)=f(x)·g(x)
h(1) if h(x)=\frac{f(x)}{g(x)}
submit question

Explanation:

Step1: Recall product - rule formula

The product - rule states that if $h(x)=f(x)\cdot g(x)$, then $h^{\prime}(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$.

Step2: Evaluate $h^{\prime}(1)$ using the product - rule

Substitute $x = 1$ into the product - rule formula. From the table, $f(1)=4$, $f^{\prime}(1)=3$, $g(1)=2$, and $g^{\prime}(1)=1$.
\[

$$\begin{align*} h^{\prime}(1)&=f^{\prime}(1)g(1)+f(1)g^{\prime}(1)\\ &=3\times2 + 4\times1\\ &=6 + 4\\ &=10 \end{align*}$$

\]

Step3: Recall quotient - rule formula

The quotient - rule states that if $h(x)=\frac{f(x)}{g(x)}$, then $h^{\prime}(x)=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g^{2}(x)}$.

Step4: Evaluate $h^{\prime}(1)$ using the quotient - rule

Substitute $x = 1$ into the quotient - rule formula. From the table, $f(1)=4$, $f^{\prime}(1)=3$, $g(1)=2$, and $g^{\prime}(1)=1$.
\[

$$\begin{align*} h^{\prime}(1)&=\frac{f^{\prime}(1)g(1)-f(1)g^{\prime}(1)}{g^{2}(1)}\\ &=\frac{3\times2-4\times1}{2^{2}}\\ &=\frac{6 - 4}{4}\\ &=\frac{2}{4}\\ &=\frac{1}{2} \end{align*}$$

\]

Answer:

$h^{\prime}(1)$ when $h(x)=f(x)\cdot g(x)$ is $10$.
$h^{\prime}(1)$ when $h(x)=\frac{f(x)}{g(x)}$ is $\frac{1}{2}$.