Question

let (f(3)=4,f^{prime}(3)=3,h(3)=5,h^{prime}(3)=2). a. if (g(z)=f(z)cdot h(z)), then (g^{prime}(3)=) b. if (g(w)=f(w)/h(w)), then (g^{prime}(3)=)

Explanation:

Step1: Recall product - rule

The product - rule states that if $G(x)=F(x)\cdot H(x)$, then $G^{\prime}(x)=F^{\prime}(x)H(x)+F(x)H^{\prime}(x)$.

Step2: Evaluate $G^{\prime}(3)$ for product - case

Substitute $x = 3$ into the product - rule formula: $G^{\prime}(3)=F^{\prime}(3)H(3)+F(3)H^{\prime}(3)$. Given $F(3) = 4,F^{\prime}(3)=3,H(3)=5,H^{\prime}(3)=2$, we have $G^{\prime}(3)=3\times5 + 4\times2=15 + 8=23$.

Step3: Recall quotient - rule

The quotient - rule states that if $G(x)=\frac{F(x)}{H(x)}$, then $G^{\prime}(x)=\frac{F^{\prime}(x)H(x)-F(x)H^{\prime}(x)}{H^{2}(x)}$.

Step4: Evaluate $G^{\prime}(3)$ for quotient - case

Substitute $x = 3$ into the quotient - rule formula: $G^{\prime}(3)=\frac{F^{\prime}(3)H(3)-F(3)H^{\prime}(3)}{H^{2}(3)}$. Plugging in $F(3) = 4,F^{\prime}(3)=3,H(3)=5,H^{\prime}(3)=2$, we get $G^{\prime}(3)=\frac{3\times5-4\times2}{5^{2}}=\frac{15 - 8}{25}=\frac{7}{25}$.

Answer:

A. $23$
B. $\frac{7}{25}$