Let $f$ and $g$ be differentiable on $I$ and suppose $f(x)g(x)=1. \forall x \in I$ Show $\frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)} =0$ Let $f$ and $g$ be differentiable on $I$ and suppose $f(x)g(x)=1$.  $ \forall x \in I$ Show $\frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)} =0$ 
This is my attempt:
Since $f$ is differentiable, then 
$ f'(c)=$$$\lim_{x\to c} \frac{f(x)-f(c)}{x-c}$$.
Since $g$ is differentiable, then  $g'(c)= $$$\lim_{x\to c}\frac{g(x)-g(c)}{x-c}$$.
Im stick here. I want to multiply $f'(c)$ by  $g(x)$ to get
$ g(x)f'(c)=$$$\lim_{x\to c} \frac{g(x)f(x)-g(x)f(c)}{x-c}$$.
and then multiply $g'(c)$ by $f(x)$ so $f(x)g'(c)= $$$\lim_{x\to c}\frac{f(x)g(x)-f(x)g(c)}{x-c}$$.
Then add the two so $f(x)g'(c)+f'(c)(g(x)$=$$\lim_{x\to c}\frac{f(x)g(x)-f(x)g(c)}{x-c} +\frac{g(x)f(x)-g(x)f(c)}{x-c}$$.
Can someone help me out. Thanks

Given $f(x)g(x)=1$, then through product rule we find $f'(x)g(x)+f(x)g'(x)=0$. Now divide by $f(x)g(x)$ and split the fraction to arrive at the desired result.
