let h(x)=f(g(x)) and p(x)=g(f(x)). use the table to compute the following derivatives.\na. h(4)\nb…

let h(x)=f(g(x)) and p(x)=g(f(x)). use the table to compute the following derivatives.\na. h(4)\nb. p(2)\n\\begin{array}{|c|c|c|c|c|}hline x&1&2&3&4\\hline f(x)&4&3&2&1\\hline f(x)& - 7& - 8& - 1& - 9\\hline g(x)&1&4&3&2\\hline g(x)&\\frac{4}{9}&\\frac{1}{9}&\\frac{7}{9}&\\frac{2}{9}\\hlineend{array}\nh(4)=\\square (simplify your answer)
Answer
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 $h^{\prime}(4)$, we substitute $x = 4$ into the formula: $h^{\prime}(4)=f^{\prime}(g(4))\cdot g^{\prime}(4)$.
Step2: Find $g(4)$ from the table
From the table, when $x = 4$, $g(4)=2$.
Step3: Find $f^{\prime}(g(4))$
Since $g(4)=2$, we need to find $f^{\prime}(2)$. From the table, $f^{\prime}(2)=-8$.
Step4: Find $g^{\prime}(4)$
From the table, when $x = 4$, $g^{\prime}(4)=\frac{2}{9}$.
Step5: Calculate $h^{\prime}(4)$
$h^{\prime}(4)=f^{\prime}(g(4))\cdot g^{\prime}(4)=-8\times\frac{2}{9}=-\frac{16}{9}$.
Answer:
$-\frac{16}{9}$
Explanation for $p^{\prime}(2)$
Step1: 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 $p^{\prime}(2)$, we substitute $x = 2$ into the formula: $p^{\prime}(2)=g^{\prime}(f(2))\cdot f^{\prime}(2)$.
Step2: Find $f(2)$ from the table
From the table, when $x = 2$, $f(2)=3$.
Step3: Find $g^{\prime}(f(2))$
Since $f(2)=3$, we need to find $g^{\prime}(3)$. From the table, $g^{\prime}(3)=\frac{7}{9}$.
Step4: Find $f^{\prime}(2)$
From the table, when $x = 2$, $f^{\prime}(2)=-8$.
Step5: Calculate $p^{\prime}(2)$
$p^{\prime}(2)=g^{\prime}(f(2))\cdot f^{\prime}(2)=\frac{7}{9}\times(-8)=-\frac{56}{9}$.
Answer for $p^{\prime}(2)$:
$-\frac{56}{9}$