Find f'(3) when f(x) = ∫ [0, 2x-1] g(t)dt and g(x) is (x-2)^2+y^2=4 (positive)

Answer 1

this is just the chain rule in reverse
If f(x) = ∫[u,v] g(t) dt then
f'(x) = g(v)*v' - g(u)*u'

now, what is y doing in there? g(x) is a function of x. and how can g(x) be an equation?

Somehow you have mangled the question.