# calculus confused

let f and g be differentiable functinos witht the following properties:

g(x)>o for all x
f(0)=1

if h(x)=f(x)g(x) and h'(x)=f(x)g'(x), then f(x)
multiple choice:
a) f'(x) b)g(x) c) e^x d)0 e)1

it can't be a or b. it also can't be c right? because if f(x) is e^x, then h'(x) would be equal to f(x)g'(x) + g(x)f'(x) but it's not.

so is it either zero or one

h(x) = f(x)g(x) --->

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

if h'(x)=f(x)g'(x), then that means that:

f'(x)g(x) = 0

Since g(x) > 0 for all x, that means that:

f'(x) = 0 for all x.

this means that f(x) is constant. Because f(0) = 1, this implies that
f(x) = 1.

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