TITLE: Suppose that $f'(x)g(x)=f(x)g'(x)$ and $g(x)\ne 0$ on (a,b). How are $f$ and $g$ related?
QUESTION [6 upvotes]: Suppose that $f'(x)g(x)=f(x)g'(x)$ and $g(x)\ne 0$ on (a,b). 
How are $f$ and $g$ related?

I posted this earlier and accidentally deleted it.
But thus far, I have:
Let $$h(x)=\frac{f(x)}{g(x)}$$
Then
$$h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$$
Since $f'(x)g(x)=f(x)g'(x)$, I propose we replace $f(x)g'(x)$ in $h'(x)$ and we get
$$h'(x)=\frac{f'(x)g(x)-f'(x)g(x)}{(g(x))^2}$$
$$=\frac{0}{(g(x))^2}$$

Is it enough, then, to say that $f$ and $g$ are constants on $(a,b)$?

REPLY [2 votes]: In your quotient rule, it must be $(-)$. Now, using the assumption, $h^\prime(x)=0$ on $(a,b)$. What can you conclude about $h$ on $(a,b)$?

REPLY [2 votes]: Remember your definition of h. You have just managed to prove that:
$$\frac{d}{dx}\frac{f(x)}{g(x)} = 0$$
This implies:
$$f(x) = Cg(x)$$
Where C is any constant.