Givens:
f(3) = 8
g(3) = 2
f'(3) = 3
g'(3) = 6
we know that:
h(x)=4f(x) + 5g(x)
we can plug in f(3) and g(3) directly
h(3) = 4f(3) + 5g(3) = 4(8) + 5(2) = 32 +10 = 42
With h'(x) we need to use the fact that derivatives are linear:
d(h(x))/dx = d/dx (4f(x) + 5g(x)) = d/dx (4f(x)) + d/dx(5g(x)) = 4f'(x) + 5g'(x)
h'(x) = 4f'(3) + 5g'(3) = 4(3) + 5(6) = 12 + 30 = 42