Question

find the following using the table below.

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

$h(1)$ if $h(x)=f(x)cdot g(x)$
$h(1)$ if $h(x)=\frac{f(x)}{g(x)}$
$h(1)$ if $h(x)=f(g(x))$
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Explanation:

Step1: Recall product - rule

If $h(x)=f(x)\cdot g(x)$, then $h^{\prime}(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$.
When $x = 1$, $h^{\prime}(1)=f^{\prime}(1)g(1)+f(1)g^{\prime}(1)$.
From the table, $f^{\prime}(1) = 3$, $g(1)=3$, $f(1)=1$, $g^{\prime}(1)=2$.
$h^{\prime}(1)=3\times3 + 1\times2=9 + 2=11$.

Step2: Recall quotient - rule

If $h(x)=\frac{f(x)}{g(x)}$, then $h^{\prime}(x)=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g^{2}(x)}$.
When $x = 1$, $h^{\prime}(1)=\frac{f^{\prime}(1)g(1)-f(1)g^{\prime}(1)}{g^{2}(1)}$.
Substitute $f^{\prime}(1) = 3$, $g(1)=3$, $f(1)=1$, $g^{\prime}(1)=2$ into the formula.
$h^{\prime}(1)=\frac{3\times3-1\times2}{3^{2}}=\frac{9 - 2}{9}=\frac{7}{9}$.

Step3: Recall chain - rule

If $h(x)=f(g(x))$, then $h^{\prime}(x)=f^{\prime}(g(x))\cdot g^{\prime}(x)$.
When $x = 1$, first find $g(1)=3$.
Then $h^{\prime}(1)=f^{\prime}(g(1))\cdot g^{\prime}(1)=f^{\prime}(3)\cdot g^{\prime}(1)$.
From the table, $f^{\prime}(3)=2$ and $g^{\prime}(1)=2$.
$h^{\prime}(1)=2\times2 = 4$.

Answer:

$h^{\prime}(1)$ when $h(x)=f(x)\cdot g(x)$ is $11$.
$h^{\prime}(1)$ when $h(x)=\frac{f(x)}{g(x)}$ is $\frac{7}{9}$.
$h^{\prime}(1)$ when $h(x)=f(g(x))$ is $4$.