The derivative operator distributes over sums, so
[tex](a f(x) + b g(x))' = a f'(x) + b g'(x)[/tex]
Then
[tex]h(x) = 5f(x) - 4g(x) \implies h'(x) = 5 f'(x) - 4 g'(x) \\\\ \implies h'(2) = 5(-2) - 4(7) = \boxed{-38}[/tex]
Use the product rule,
[tex](f(x)g(x))' = f'(x) g(x) + f(x) g'(x)[/tex]
Then
[tex]h(x) = f(x) g(x) \implies h'(x) = f'(x) g(x) + f(x) g'(x) \\\\ \implies h'(2) = (-2) (4) + (-3) (7) = \boxed{-29}[/tex]
Use the quotient rule,
[tex]\left(\dfrac{f(x)}{g(x)}\right)' = \dfrac{f'(x) g(x) - f(x) g'(x)}{g(x)^2}[/tex]
Then
[tex]h(x) = \dfrac{f(x)}{g(x)} \implies h'(x) = \dfrac{f'(x) g(x) - f(x) g'(x)}{g(x)^2} \\\\ \implies h'(2) = \dfrac{(-2)(4) - (-3)(7)}{4^2} = \boxed{\dfrac{13}{16}}[/tex]
Use the quotient rule again.
[tex]h(x) = \dfrac{g(x)}{1 + f(x)} \implies h'(x) = \dfrac{g'(x) (1+f(x)) - g(x) f'(x)}{(1+f(x))^2} \\\\ \implies h'(2) = \dfrac{7(1 - 3) - 4 (-2)}{(1 - 3)^2} = -\dfrac64 = \boxed{-\dfrac32}[/tex]
