Question

let h(x)=f(g(x)) and p(x)=g(f(x)). use the table to compute the following derivatives. a. h(1) b. p(3)

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

Explanation:

Step1: Apply 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 $h^{\prime}(1)$, first find $g(1)$ from the table. When $x = 1$, $g(1)=1$. Then find $f^{\prime}(g(1))=f^{\prime}(1)$ and $g^{\prime}(1)$. From the table, $f^{\prime}(1)=-4$ and $g^{\prime}(1)=\frac{1}{5}$. So $h^{\prime}(1)=f^{\prime}(g(1))\cdot g^{\prime}(1)=-4\times\frac{1}{5}=-\frac{4}{5}$.

Step2: Apply 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 $p^{\prime}(3)$, first find $f(3)$ from the table. When $x = 3$, $f(3)=2$. Then find $g^{\prime}(f(3))=g^{\prime}(2)$ and $f^{\prime}(3)$. From the table, $g^{\prime}(2)=\frac{3}{5}$ and $f^{\prime}(3)=-3$. So $p^{\prime}(3)=g^{\prime}(f(3))\cdot f^{\prime}(3)=\frac{3}{5}\times(-3)=-\frac{9}{5}$.

Answer:

$h^{\prime}(1)=-\frac{4}{5}$
$p^{\prime}(3)=-\frac{9}{5}$