TITLE: Where is the mistake in my 'derivation' of $(uv)' = u'v$?
QUESTION [9 upvotes]: 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?

REPLY [2 votes]: 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).