Where is the mistake in my 'derivation' of $(uv)' = u'v$? So I tried to derive the product rule without adding $f(x)g(x+\Delta x)-f(x)g(x+\Delta x)$ as we used to. Instead I started deriving it directly and ran into a strange conclusion that $(uv)'=u'v$. The derivation looks like this: 
\begin{align}
(uv)' & = \lim_{\Delta x\to0} \frac{u(x+\Delta x)v(x+\Delta x)-u(x)v(x)}{\Delta x} \\
& = \lim_{\Delta x\to0} \frac{u(x+\Delta x)v(x+\Delta x)}{\Delta x}-\lim_{\Delta x\to0} \frac{u(x)v(x)}{\Delta x} \\
& = \lim_{\Delta x\to0} \frac{u(x+\Delta x)}{\Delta x}\lim_{\Delta x\to0} v(x+\Delta x)-\lim_{\Delta x\to0} \frac{u(x)}{\Delta x}\lim_{\Delta x\to0} v(x) \\
& = \lim_{\Delta x\to0} \frac{u(x+\Delta x)}{\Delta x}v(x)-\lim_{\Delta x\to0} \frac{u(x)}{\Delta x}v(x) \\
& = v(x)(\lim_{\Delta x\to0} \frac{u(x+\Delta x)}{\Delta x}-\lim_{\Delta x\to0} \frac{u(x)}{\Delta x}) \\
& = v(x)\lim_{\Delta x\to0} \frac{u(x+\Delta x)-u(x)}{\Delta x} \\
& = u'v
\end{align}
Apparently there is a mistake somewhere, but I can't figure out where exactly. Any ideas?
 A: I find your notation very cumbersome. Now, taking into account that differentiability implies continuity, we can write
$$\frac{f(x)g(x)-f(x_0)g(x_0)}{x-x_0}=\frac{f(x)g(x)-f(x)g(x_0)+f(x)g(x_0)-f(x_0)g(x_0)}{x-x_0}=$$
$$=f(x)\frac{g(x)-g(x_0)}{x-x_0}+\frac{f(x)-f(x_0)}{x-x_0}g(x)\xrightarrow[x\to x_0]{}f(x_0)g'(x_0)+f'(x_0)g(x_0)$$
Complete details. The above is the easiest proof I know of the product rule for derivative. The trick in the  first step is also used in general limits.
A: This has been pointed out in comments already, but the error in the proof is introduced in this step:
$$ \lim_{\Delta x\to0} \frac{u(x+\Delta x)v(x+\Delta x)}{\Delta x}-\lim_{\Delta x\to0} \frac{u(x)v(x)}{\Delta x}.$$
Neither of these limits exists (in both cases the quantity goes to infinity),
so the difference is indeterminate. Once you start manipulating such a thing as if such limits were valid, it is likely to come out to any number of wrong answers. By inserting a couple more steps afterward, you could reduce the entire expression to zero.

Taking the idea that a $u'$ has to do with the change in $u$ over the change in $x$, write $\Delta u = u(x+\Delta x) - u(x).$
Likewise write $\Delta v = v(x+\Delta x) - v(x).$
That is, $u(x+\Delta x) = u(x) + \Delta u$ and $v(x+\Delta x) = v(x) + \Delta v.$
Now plug this into your first formula and plod along without knowing where we're going until we get there.
\begin{align}
(uv)' 
&= \lim_{\Delta x\to0} \frac{u(x + \Delta x)v(x + \Delta x) - u(x)v(x)}{\Delta x} \\
&= \lim_{\Delta x\to0} \frac{(u(x) + \Delta u)(v(x) + \Delta v) - u(x)v(x)}{\Delta x} \\
&= \lim_{\Delta x\to0} \frac{(u(x)v(x) + (\Delta u)v(x) + u(x)(\Delta v)
                               + (\Delta u)(\Delta v)) - u(x)v(x)}{\Delta x} \\
&= \lim_{\Delta x\to0} \frac{(\Delta u)v(x) + u(x)(\Delta v)
                               + (\Delta u)(\Delta v)}{\Delta x} \\
\end{align}
Now observe that
\begin{align}
\lim_{\Delta x\to0} \frac{\Delta u}{\Delta x} 
&= \lim_{\Delta x\to0} \frac{u(x+\Delta x) - u(x)}{\Delta x} = u'(x), \\
\lim_{\Delta x\to0} \frac{\Delta v}{\Delta x} 
&= \lim_{\Delta x\to0} \frac{v(x+\Delta x) - v(x)}{\Delta x} = v'(x), \\
\lim_{\Delta x\to0} \frac{(\Delta u)(\Delta v)}{\Delta x} 
&= \left(\lim_{\Delta x\to0} \frac{\Delta u}{\Delta x}\right)
   \left(\lim_{\Delta x\to0} \Delta v\right) = 0
\end{align}
since $\lim_{\Delta x\to0} \Delta v = 0.$
Putting all of this together,
\begin{align}
(uv)' 
&= v(x)\lim_{\Delta x\to0} \frac{\Delta u}{\Delta x}
  + u(x)\lim_{\Delta x\to0} \frac{\Delta v}{\Delta x}
  + \lim_{\Delta x\to0} \frac{(\Delta u)(\Delta v)}{\Delta x} \\
&= v(x) u'(x) + u(x) v'(x) + 0 \\
&= u' v + u v'.
\end{align}
No tricks with adding and subtracting a mysterious term or other great inspiration, just distribution of multiplication over addition and beating on the monster until it's dead.
Personally I like the more inspired solutions, but sometimes when you're stuck for inspiration you can just push your way through.
A: With regard to where you went wrong in your original posting, it is in the second line. You have writen a limit of the form $\infty -\infty$ which is indeterminate. Remember, $x$ is fixed, so $u(x), v(x)$ are also fixed, therefore, the factor  $\frac{u(x+\Delta x)}{\Delta x}$ of the first term in line 4 and $\frac{u(x)}{\Delta x}$ of the second term of line 4 have infinite limits as $\Delta x \rightarrow 0$. But really, you already had this error present from line 2. 

I want to mention another way to 'arrive' at the product rule using logarithms. It is important to recognize that this is not a proof, except when the functions $u,v$ are strictly positive. However, it is a convenient symbolic way to recall the product rule (a useful mnemonic).  If $y=uv$ then 
\begin{align} \log{y}&=\log{uv}&\\
&=\log{u} +\log{v}\end{align}
Hence, 
$\frac{y'}{y}=\frac{u'}{u}+\frac{v'}{v}$
Now, multiply by $y$ and you get $y'=u'v+v'u$. Again, this is not a proof of the general product rule, but other posters have already given satisfactory explanations using the difference quotient definition. 
I personally like this method because it also works to produce a formula for the derivative of a product of an arbitrary finite number of functions (provided that all of these functions are strictly positive!). 
If $y=\prod_{j=1}^n f_j$ then 
\begin{align} \log{y}&=\log{\prod_{j=1}^n f_j}&\\
&= \sum_{j=1}^n \log{f_j}\end{align}
therefore, 
\begin{align} \frac{y'}{y}= \sum_{j=1}^n \frac{f_j'}{f_j} \end{align}
hence 
\begin{align} y'&=\sum_{j=1}^n \frac{yf_j'}{f_j}\\
&= \sum_{j=1}^n \left(f_j'(x)\prod_{\stackrel{1\leq i\leq n}{i\neq j}}f_j(x)\right) \end{align}
A: As others have pointed out, and to make @Pakk happy, the error is in the second line. Now I'm going to go on and answer the question you didn't ask, but probably should have: 

How could I, too, learn to detect where the error is?

Well, you know that $(uv)' = u'v + u v'$, right? But you've "proved" that $(uv)' = u'v$. So if you can find functions with $uv' \ne 0$, the two right-hand-sides will be different. There are a lot of functions $u$ and $v$ with $uv' \ne 0$ of course, but picking a really simple pair will make things especially easy to de-bug. So let's pick $u$ to be a constant: $u(x) = 1$, and $v$ to be something whose derivative is constant: $v(x) = x$. Now let's look at your "proof" for those two functions. I'm just going to substitute in these particular values for $u$ and $v$ (or their derivatives) wherever they occur in your sequence of limits:
\begin{align}
(uv)' & = \lim_{\Delta x\to0} \frac{u(x+\Delta x)v(x+\Delta x)-u(x)v(x)}{\Delta x} \\
& = \lim_{\Delta x\to0} \frac{u(x+\Delta x)v(x+\Delta x)}{\Delta x}-\lim_{\Delta x\to0} \frac{u(x)v(x)}{\Delta x} \\
& = \lim_{\Delta x\to0} \frac{u(x+\Delta x)}{\Delta x}\lim_{\Delta x\to0} v(x+\Delta x)-\lim_{\Delta x\to0} \frac{u(x)}{\Delta x}\lim_{\Delta x\to0} v(x) \\
& = \lim_{\Delta x\to0} \frac{u(x+\Delta x)}{\Delta x}v(x)-\lim_{\Delta x\to0} \frac{u(x)}{\Delta x}v(x) \\
& = v(x)(\lim_{\Delta x\to0} \frac{u(x+\Delta x)}{\Delta x}-\lim_{\Delta x\to0} \frac{u(x)}{\Delta x}) \\
& = v(x)\lim_{\Delta x\to0} \frac{u(x+\Delta x)-u(x)}{\Delta x} \\
& = u'v
\end{align}
becomes (when we evaluate at $x = 1$, which is as good a place as any):
\begin{align}
(uv)'(1) & = \lim_{\Delta x\to0} \frac{u(1+\Delta x)v(1+\Delta x)-u(1)v(1)}{\Delta x} \\
(uv)'(1) & = \lim_{\Delta x\to0} \frac{1\cdot (1+\Delta x)-1 \cdot 1}{\Delta x} \\
& = \lim_{\Delta x\to0} \frac{1\cdot(1+\Delta x)}{\Delta x}-\lim_{\Delta x\to0} \frac{1\cdot 1}{\Delta x} \\
\ldots
\end{align}
and now suddenly the "bug" jumps out at you: the thing on the second line, which is $1$, is not at all what you've got on the third line, which is a difference of two limits that don't exist. 
By the way, a different bug, namely that the left-hand side of what you wrote is a function, while the right-hand side, if the limit exists, is a real number, also becomes pretty obvious when you try to check things this way. The LHS of your sequence of equalities should have been $(uv)'(x)$, not $(uv)'$. I swept that error under the rug with my "when we evaluate at $x = 1$..." remark so that I could get to demonstrating the more useful skill.
Short summary: If you've proved something general that you suspect is wrong, find a specific example where it appears to produce the wrong answer, and trace that example through your supposed "proof" to find the error (if there is one). 
A: Remember you may only use the additive and multiplicative rules for limits when you can justify that those limits do have finite convergence.   So we just need to arrange the limit definition of $[uv]'$ into that for $u\cdot v'+u'\cdot v$ (or vice versa).   This is fairly straight forward, as long as you recognise that: $u(x)=\lim\limits_{h\to 0}u(x{+}h)$.
$$\begin{align}[uv'+u'v](x)&=u(x)\left(\lim_{h\to 0}\dfrac{v(x{+}h)-v(x)}{h}\right)+\left(\lim_{h\to 0}\dfrac{u(x{+}h)-u(x)}{h}\right)v(x)
\\[1ex]&= \left(\lim_{h\to 0}u(x{+}h)\right)\left(\lim_{h\to 0}\dfrac{v(x{+}h)-v(x)}{h}\right)+\left(\lim_{h\to 0}\dfrac{u(x{+}h)-u(x)}{h}\right)v(x)
\\[1ex]&=\lim_{h\to 0}\dfrac{u(x{+}h)\left(v(x{+}h)-v(x)\right)+\left(u(x{+}h)-u(x)\right)v(x)}{h}\\[1ex]&=\lim_{h\to x}\dfrac{u(x{+}h)\,v(x{+}h)-u(x{+}h)\,v(x)+u(x{+}h)\,v(x)-u(x)\,v(x)}{h}\\[1ex]&=\lim_{h\to 0}\dfrac{u(x{+}h)\,v(x{+}h)-u(x)\,v(x)}{h}\\[3ex][uv'+u'v](x)&= [uv]'(x)\end{align}$$
