Question

let h(x)=f(g(x)) and p(x)=g(f(x)). use the table to compute the following derivatives.
a. h(4)
b. p(2)

x	1	2	3	4
f(x)	4	1	3	2
f(x)	-6	-7	-9	-2
g(x)	2	4	1	3
g(x)	5/7	1/7	4/7	6/7

Explanation:

Step1: Recall chain - rule for $h(x)$

The chain - rule states that if $h(x)=f(g(x))$, then $h^{\prime}(x)=f^{\prime}(g(x))\cdot g^{\prime}(x)$. To find $h^{\prime}(4)$, we substitute $x = 4$ into the formula: $h^{\prime}(4)=f^{\prime}(g(4))\cdot g^{\prime}(4)$.

Step2: Find $g(4)$ and $g^{\prime}(4)$ from the table

From the table, when $x = 4$, $g(4)=3$ and $g^{\prime}(4)=\frac{6}{7}$.

Step3: Find $f^{\prime}(g(4))$

Since $g(4)=3$, we find $f^{\prime}(3)$ from the table. When $x = 3$, $f^{\prime}(3)=-9$.

Step4: Calculate $h^{\prime}(4)$

$h^{\prime}(4)=f^{\prime}(g(4))\cdot g^{\prime}(4)=(-9)\times\frac{6}{7}=-\frac{54}{7}$.

Step5: Recall chain - rule for $p(x)$

The chain - rule states that if $p(x)=g(f(x))$, then $p^{\prime}(x)=g^{\prime}(f(x))\cdot f^{\prime}(x)$. To find $p^{\prime}(2)$, we substitute $x = 2$ into the formula: $p^{\prime}(2)=g^{\prime}(f(2))\cdot f^{\prime}(2)$.

Step6: Find $f(2)$ and $f^{\prime}(2)$ from the table

From the table, when $x = 2$, $f(2)=1$ and $f^{\prime}(2)=-7$.

Step7: Find $g^{\prime}(f(2))$

Since $f(2)=1$, we find $g^{\prime}(1)$ from the table. When $x = 1$, $g^{\prime}(1)=\frac{5}{7}$.

Step8: Calculate $p^{\prime}(2)$

$p^{\prime}(2)=g^{\prime}(f(2))\cdot f^{\prime}(2)=\frac{5}{7}\times(-7)=-5$.

Answer:

$h^{\prime}(4)=-\frac{54}{7}$
$p^{\prime}(2)=-5$