# 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?

• Most of the limits you have written during your working do not exist due to division by zero. Hence this kind of invalidates the working. – Peter Foreman Jul 17 at 22:12
• $\lim_{\Delta x \rightarrow 0} \frac{v(x)u(x)}{\Delta x}$ is certainly not defined in general – Dayton Jul 17 at 22:13
• We can't say that $\lim_{x \to a} f(x) - g(x) = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)$ if either $f$ or $g$ do not have a limit as $x$ approaches $a$. – littleO Jul 17 at 22:18
• In a 'where is my mistake?'-question like this, why do so many answers ignore that question and just give alternative proofs? Am I the only one bothered by this? – Pakk Jul 18 at 6:53
• @TeiReiDa, no need to apologize, I think it is a good and clear question, and I don't know why other people interpret it as "please ignore all the work I did and just give me an alternative way to proof this". – Pakk Jul 18 at 10:15

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}

• @Pakk: The "easy way" given in this answer is simply wrong. It fails when $u$ is the constant zero function, among many other problems. – user21820 Jul 18 at 7:15
• @user218120: The answer starts at "With regard to where you went wrong", everything before that is irrelevant. – Pakk Jul 18 at 7:32
• @Pakk: Yes the second half attempts to explain what went wrong in the asker's question, which is a good thing. But I don't consider a post that is half wrong to be a correct answer, as readers are typically incapable of identifying what is wrong. I'll wait for user140776 to remove the incorrect first half. – user21820 Jul 18 at 8:20
• It is the only answer that gives an answer to the real question ("it is in the second line"), and it gives a correct answer to that. I agree that the answer would improve if the first part is removed, because it is irrelevant, and also wrong. But this answer is better than the other 'anwers' that don't answer the question. – Pakk Jul 18 at 10:22
• That $\log$ thing may be formally wrong but it is also an excellent mnemonic, especially when you apply the same principle to $y=\frac u v$, which is hell to memorise otherwise. – Martin Kochanski Jul 18 at 11:40

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.

• Yes, I know. But for some reason this addition-subtraction step wasn't the first that came to my mind when I decided to do this derivation as a warmup. For me in most cases it's unclear what exactly should I add and subtract to simplify the expression, so usually I prefer to evade such methods. – TeiReiDa Jul 17 at 23:05
• That preference is okay, but sometimes it is unevadable. After all, you need to arrange the limit definition of $[uv]'$ into that for $u\cdot v'+u'\cdot v$. $$\lim_{h\to 0}\tfrac{[uv](x+h)-[uv](x)}h=u(x)\lim_{h\to 0} \tfrac{v(x+h)-v(x)}h+v(x)\lim_{h\to 0}\tfrac{u(x+h)-u(x)}h$$ [The trick is recognising $u(x)=\lim\limits_{h\to 0}u(x{+}h)$] – Graham Kemp Jul 17 at 23:24

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.

• This would be a great answer to the question "How to derive the product rule?". Sadly, a different question was asked, that is not answered in this post. – Pakk Jul 18 at 10:26
• @Pakk At the time I posted, I was aware that the question "where is the mistake" had already been answered, so I chose not to repeat that part of the answer. I merely answered the next obvious question, "How could I have derived the product rule directly without adding $f(x)g(x+\Delta x)-f(x)g(x+\Delta x)$?" But since the accepted answer turns out only partially correct, I've added to this one. – David K Jul 18 at 11:18
• And now, this answer is the best answer. I'm happy. Have a great day! – Pakk Jul 18 at 11:43
• Typographical error: $$\lim_{\Delta x \to 0} \frac{\Delta v}{\Delta x} = lim_{\Delta x \to 0} \frac{v(x + \Delta x) - v(x)}{\Delta x} = \color{red}{v}'(x)$$ – N. F. Taussig Jul 19 at 9:01
• @N.F.Taussig Thanks for spotting that. Fixed. – David K Jul 19 at 11:10

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).

• very nice and immensely useful last paragraph! +1 – peek-a-boo Jul 18 at 12:10
• I am even happier than I was before, this day is the highlight of my life! – Pakk Jul 18 at 19:37
• While I was being silly in my answer, I actually share your frustration about people answering the wrong question. – John Hughes Jul 18 at 19:57
• In highschool I had a bad teacher, who never explained what was wrong about a failed attempt, but just showef his preferred method. I saw classmates passing the tests because they did what he told, while never understanding why they should do it in that way, and they understandably started to hate mathematics. They were trained to not think about it, and just memorize what the teacher said. That is the source of my frustration when the attempt is ignored, and another method is given. – Pakk Jul 20 at 17:57

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}