On [a, b], if f ¢ (x) = g¢ (x) then f
(x) = g(x) + C
First, a little rewrite:
f ¢ (x) = g¢ (x)
f ¢ (x) – g¢ (x) = 0
f (x) = g(x) + C
f (x) – g(x) = C
So define h(x) = f (x) – g(x). Then it will be
sufficient to show that if h¢ (x) = 0, then h(x) = C.
Notice that h must be
differentiable on (a, b).