Help with understanding proof of the product rule $F(x)=f(x)g(x)$  
$\lim_{h \to 0}\dfrac{f(x+h)g(x+h)-f(x)g(x)}{h}$  
Then, the notes I'm reading say: "the numerator $f(x+h)g(x+h)-f(x)g(x)$ is a difference that involves $x$ changing to $x+h$ for both $f$ and $g$. We need to isolate the change for $f$ from the change for $g$.   
I don't understand this concept and this explanation, can you please elaborate? Why do we have to isolate and why can't we have $x$ changing to $x+h$ for both $f$ and $g$? 
Then the notes rewrite the numerator $f(x+h)g(x+h)-f(x)g(x)$ as:
$(f(x+h)-f(x))g(x)+f(x)(g(x+h)-g(x))+(f(x+h)-f(x))(g(x+h)-g(x))$  
and I don't understand this either.   
I'm trying to look at the product rule proof geometrically but I'm not understanding what we are doing in the geometric proof.   
Thank you. 
 A: I don't know the idea of the rewritten numerator. Proving $F'(x)=f'(x)g(x) + f(x)g'(x)$ geometrically, you can use
\begin{align*}
\Delta F&=\Delta F+0\\
&=f(x+h)g(x+h)\color{red}{-f(x)g(x+h)}-f(x)g(x)+\color{red}{f(x)g(x+h)}\\
&=g(x+h)\left [ f(x+h)-f(x) \right ]+f(x)\left [ g(x+h)-g(x) \right ]\\
&=g(x+h)\Delta f+f(x)\Delta g.
\end{align*}
The expression $0=f(x)g(x+h)-f(x)g(x+h)$ is used as a "trick" to simplify the proof. You can also use $0=g(x)f(x+h) - g(x)f(x+h)$.
Hence, 
$$\lim_{h\to 0}\frac{\Delta F}{h}=g(x)\lim_{h\to 0}\frac{\Delta f}{h}+f(x)\lim_{h\to 0}\frac{\Delta g}{h}=F'(x).$$
A: Start with a rectangle with base $f(x)$ and height $g(x)$, and enlarge the base by a distance $f(x+h)-f(x)$ and the height by a distance $g(x+h)-g(x)$ to get a new rectangle with base $f(x+h)$ and $g(x+h)$.  
Then the area inside the new rectangle and outside the original rectangle can be cut into 3 smaller rectangles, with areas $A_1=(f(x+h)-f(x)(g(x))$, $\;\;\;$
$A_2=(f(x))(g(x+h)-g(x)$, and $\hspace{.68 in}A_3=(f(x+h)-f(x))(g(x+h)-g(x))$, 
so the sum of these 3 areas represents the difference in the areas of the old and new rectangles:  $A_1+A_2+A_3=(f(x+h)-f(x))(g(x+h)-g(x))$.
[As Andre Nicolas points out, though, this is not the usual way the numerator is rewritten to prove the Product Rule.]
A: I think this is not the thing that you exactly need.
But I like to use logarithms ( I think you are familiar with derivatives of logarithms).
Let $$y=f(x)g(x),$$
then $$\ln y=\ln f(x)+\ln g(x).$$
$$\frac{1}{y}\frac{dy}{dx}=\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}$$ 
$$\frac{dy}{dx}=f'(x)g(x)+g'(x)f(x).$$
This works not only for this but also for $$y=\frac{f(x)}{g(x)}$$ and $$y=(f(x))^{g(x)}.$$  
