Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 64

Find $h'(2)$, given that $f(2) = -3$, $g(2) = 4$, $f'(2) = -2$, and $g'(2) = 7$

$\begin{equation} \begin{aligned} \text{a.) } h(x) &= 5f(x) - 4g(x) &&& \text{b.) } h(x) &= f(x) g(x)\\ \text{c.) } h(x) & = \frac{f(x)}{g(x)} &&& \text{d.) } h(x) &= \frac{g(x)}{1+f(x)} \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} \text{a.) } h(x) & = 5 f(x) - 4g(x)\\ \\ h'(2)& = 5[f'(2)] - 4 [g'(2)]\\ \\ h'(2)& = 5(-2) -4(7)\\ \\ h'(2)& = - 38 \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} \text{b.) } h(x) & = f(x)4g(x)\\ \\ h'(2)& = f'(2) [g(2)] + g'(2) [f(2)]\\ \\ h'(2)& = -2(4) + 7(-3)\\ \\ h'(2)& = - 29 \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} \text{c.) } h(x) & = \frac{f(x)}{g(x)}\\ \\ h'(2)& = \frac{g(2)[f'(2)]-f(2)[g'(2)]}{[g(2)]^2}\\ \\ h'(2)& = \frac{4(-2)-(-3)(7)}{(4)^2}\\ \\ h'(2)& = \frac{13}{16} \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} \text{d.) } h(x) & = \frac{g(x)}{1+f(x)}\\ \\ h'(2)& = \frac{[1+f(2)]g'(2)-g(2)[0+f'(2)]}{[1+f(2)]^2}\\ \\ h'(2)& = \frac{[1+(-3)]7-4[0+(-2)]}{[1+(-3)]^2}\\ \\ h'(2)& = \frac{-3}{2} \end{aligned} \end{equation}$