Dark Blue Magic for Product Rule

If $h(x)=f(x)g(x)$ , what is $h'(x)$ ? (help with notation in Appendix A)

(If at any point you get stuck, go to Appendix B)

Our goal is to test different examples, looking for solutions that we think might work until we get down to only one possible candidate solution that fits all tests, and then test it two more times. What actually happened–it wasn’t a matter of thinking something would work. It was trying something to see what would happen, and proving what was wrong with it.

If the equation we write for the Product Rule is correct, it will work for any pairing of f(x) and g(x). We could do something like test $f(x)=x^2$ and $g(x)=y^3$ , and we would know the answer because $h(x)=y^5$ and we know the derivative of a monomial, so $h'(x)=5y^4$ .

We have yet another option–we can choose for one of the two functions to be “1”. This is “tricky”, sometimes it forces the correct answer to step forward fairly quickly, and sometimes it fails miserably.

Let $f(x)=2x$ and $g(x)=1$ .

Our first idea from Black Magic to use is the idea that the derivative, $h'(x)$ , should have pieces or things coming from the original function, $h(x)$ . Does it make sense that the derivative of the product of two functions might have the derivatives of those two functions in it?

$f'(x)=2$
$g'(x)=0$

We entertain the idea, that maybe, “the derivative of a product is the product of the derivatives”, so we test the following:

[/latex]f'(x) g'(x) = 0[/latex]

That rules that out. Next, we consider, well, we do notice that [/latex]f'(x)[/latex] is equal to the answer we want. But that doesn’t make sense, how could the derivative of the product of “f” and “g” have nothing coming from g?

We think about it, we could switch “f” and “g” and the product would still be the same, therefore the answer has to be built in a way that switching “f” and “g” won’t change the answer. That’s commutation and it takes us to things like “f*g = g*f” and “(f+g)=(g+f)”.

Well, looking at these we might say “we just failed with f’*g’, what about $f'(x)+g'(x)$ ?

[/latex]f'(x) + g'(x) = 2 + 0[/latex]

This gives us the right answer, but we are not done. We did not use rigorous mathematical methods to prove the answer. Instead, we did guesswork to try to build the right answer. You have the right, right now, to say, “I feel I am halfway or a third of the way to having a hypothesis.”

We don’t tell anyone, except maybe a few best friends who can share a secret.

We then try it with

$f(x) = x^2$
$g(x) = x^3$
$h(x) = f(x)g(x) = x^5$

We can calculate the derivative:

$h'(x) = 5x^4$

Now for the derivatives:

$f'(x) = 2x$
$g'(x) = 3x^2$

Anything contributing to the answer, needs to be a term with $x^4$ in it. For the four pieces we have, two factor multiplications are limited to (2x) with (x^3) and (3x^2) with (x^2). What happens if we sum these?

$(2x)(x^3) + (3x^2)(x^2) = (2x^4) + (3x^4) = 5x^4$

The summation gives us the right answer. I STRESS that we did not know this was going to work out, there was no hidden knowledge/conspiracy to tell us this was the road to take, we knew we were there only after we saw the $5x^4$ .

$f'(x) g(x) + g'(x) f(x)$

If we modify our notation and write this as, f’ g + g’ f, does that help you to see the Commutation? Can you see that rewriting this, switching $\text{[math]}$ with $\text{[math]}$ , doesn’t change the statement? Note that we’re utilizing commutation of multiplication and commutation of addition.

We can go back and test this with the first problem, and it will work. You’ll be thankful for that zero cancelling that second term when the first term held the correct answer!

Suggestion. We call this a hypothesis at this point. Maybe now you have the confidence to release this to an entire school. Suddenly you are famous and have a fan club. Hundreds of people write back they have played with it, and so far, so good. Now tell people you want to call it a theory. I’m not sure what to tell you if your question is “when will they make it a law?”

Appendix A – Notation for Derivatives

If you apply the equation to get get several point slopes for points on the equation $f(x)=x^2$ , you’ll soon notice that your points go on the line we get from 2x, thus the derivative is 2x.

$f(x)=x^2$

Below are four different ways to show the derivative:

$f'(x) = 2x$
$\dfrac {d}{dx} x^2 = 2x$
$\dfrac {df}{dx} = 2x$
$\dot{f} = 2x$

Appendix B – If you are new to this, you shouldn’t feel that you are supposed to be able to guess what needs to be done, and it should feel weird at first. If I succeed at what I want, it will be that I give you enough examples, that, over time, you start getting a feel for what can be done.

This topic is on the edge, it has no right to speak to you regarding your self confidence. If necessary, tell it to get back in its cage!

Appendix C – This appendix is provided as a bonus feature at no additional charge to your account. You could read through a document like this and it might take five to ten minutes. The document probably won’t tell you “oh, we had this question and someone worked on it for 40 years before one night they dreamed a cougar ran up to them and put its paws on their shoulders and licked their face and when they woke up they were startled into thinking something that led to the answer.” It’s so much shorter and prestigious to say “I used logic to get from my question to my answer, and here’s how.”

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