Need help with Calculus problem

Let f and g be differentiable functions with the following properties:

(i) g(x)>0 for all x,
(ii) f(0)=1

if h(x)=f(x)g(x) and h'(x)=f(x)g'(x), then f(x)=

(a) f'(x)
(b) g(x)
(c) e^x
(d) 0
(e) 1


Explain step by step please. And what exactly is the question asking? 10 points.

Normally for:
h(x) = f(x)g(x)
h'(x) = f(x)g'(x)+ f'(x)g(x) => notice no second term? now since g(x) > 0 for all x, then f '(x) must be zero, but f(0) = 1 hence:
f(x) = 1
e => Answer

h '(x) = f(x)* g '(x) + g(x)* f '(x)

This means that g(x) * f '(x) = 0 and by examining the problem, you can see that g(x) > 0 for all x, therefore f '(x) = 0

Therefore, f(x) must be some constant, so f(x) = 1 since f(0) = 1.

By using the product rule
h'(x) = f'(x)g(x) + f(x)g'(x)

By using the given information, we find that f'(x)g(x) = 0
That means that f'(x) = 0 since g(x) > 0 for all x
For f'(x) = 0 means that f(x) = C, a constant
Since f(0) = 1 then f(x) = 1

answer e