The logic in the proof of the product rule http://planetmath.org/proofofproductrule
Here is a proof of the product rule. I understand how it all works, but I want to know the reasoning behind why whoever came up with this chose to add a $f(x+h)g(x) - f(x+h)g(x)$ in the numerator. I see it works out, but how can someone just think I should put this in to get a result that allows for a shortcut of taking the derivative of a product? Maybe it just takes a very very long time working with mathematics to be able to see such connections and that is all there is to it.  
 A: Look at the original expression
$$f(x+h)g(x+h)-f(x)g(x).$$
Something very complicated is going on: we are trying to find how much the product changes when both factors change at once. Don't you think it might just possibly be easier to see what happens if we change one factor at a time? So we start with
$$f(x)g(x)\tag1$$
and first change $f(x)$ to $f(x+h)$
$$f(x+h)g(x)\tag2$$
and after that change $g(x)$ to $g(x+h)$
$$f(x+h)g(x+h).\tag3$$
The increment from (1) to (2) is
$$f(x+h)g(x)-f(x)g(x),$$
and the increment from (2) to (3) is
$$f(x+h)g(x+h)-f(x+h)g(x),$$
and the total increment from (1) to (3) is
$$[f(x+h)g(x+h)-f(x+h)g(x)]+[f(x+h)g(x)-f(x)g(x)].$$
The idea "change one variable at a time, rather than all at once" may be clever, but not inhumanly clever.
A: The reason that they put it in, as you probably already know, is so that it we can easily factor with it.
As for why someone thought of this, notice that if we did not add this in, there would be no connection between $f(x),g(x)$. When we add this connection it, it easily allows for factoring and simplification. 
It's not that it takes very long to do it without it, it's that you cannot do it without it. You need that "connection", in order for the proof to simplify down and make sense.
A: The logic of the product rule is to capture the intuition behind this image.
Visual Proof of Product Rule
A: This may not be what you're after, but there is a slightly different way of looking at a derivative which makes the proof of the product rule very natural. Let $f$ be a function which we want to differentiate at $c$. Say that $f$ is differentiable at $c$ if there is some continuous function $f_c$ such that $$f(x) = f(c) + (x - c) f_c(x)$$ and if such a function $f_c$ exists, say that the derivative of $f$ at $c$ is $f'(c) := f_c(c)$. Note that this agrees with the limit definition just by rearranging the equation, but this definition makes a lot of derivations much easier, and each step in the derivation is natural.
The rest of the answer is a proof that if $f$ and $g$ are differentiable at $c$, then their product $h(x) = f(x)g(x)$ has the derivative $h'(c) = f'(c) g(c) + f(c) g'(c)$ at $c$.
Now, suppose both $f$ and $g$ are differentiable at $c$. That means we have continuous functions $f_c$ and $g_c$ such that
$$\begin{aligned}
f(x) &= f(c) + (x - c) f_c(x) \\
g(x) &= g(c) + (x - c) g_c(x)
\end{aligned}$$
Multiplying these gives
$$f(x) g(x) = f(c) g(c) + (x - c)\left[ f_c(x) g(c) + f(c) g_c(x) + (x - c) f_c(x) g_c(x) \right] $$
So we've found that the function $h(x) = f(x) g(x)$ may be written $$h(x) = h(c) + (x - c) h_c(x)$$
where $h_c(x) = f_c(x) g(c) + f(c) g_c(x) + (x - c) f_c(x) g_c(x)$. So the derivative of $h$ at $c$ should be $$h'(c) := h_c(c) = f_c(c) g(c) + f(c) g_c(c) = f'(c) g(c) + f(c) g'(c)$$
A: If you find the trick of adding and subtracting $f(x+h) g(x) $ so surprising it is better to use a more direct approach. Since we have $$f'(x) =\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ we have $$f(x+h) =f(x) +h\{f'(x)+\rho\} $$ where $\rho\to 0$ with $h$. Similarly $$g(x+h) =g(x) +h\{g'(x)+\rho'\} $$ where $\rho'\to 0$.
Now we can see that $$f(x+h) g(x+h) =f(x)g(x) + h\{f(x) g'(x) +g(x) f'(x) \} + h\{\rho'f(x)+\rho g(x)\} +h^2\{(f'(x)+\rho)(g'(x)+\rho')\}$$ and then $$\frac{f(x+h) g(x+h) - f(x) g(x)} {h} =f(x) g'(x) +g(x) f'(x) + \rho'g(x) +\rho f(x) +h\{(f'(x) +\rho) (g'(x) +\rho') \}$$ Now letting $h\to 0$ we get the desired result. 
