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This is a very quick question, and it might be pretty basic. But as a preface, I plan to dive into the actual proofs behind the derivative rules after this post, but I would like to see if my intuition is correct here first!

I was toying around with derivative rules, and I had the thought that, no matter what form your function is in, you will be performing the product rule!

For example, if you want to take a derivative of the function $f(x) = x^2\times sin(x)$, you would apply the product rule and get $\frac{d}{dx} f(x) = 2x\times sin(x) + x^2\times cos(x)$

This is an obvious use of the product rule, but I think that there is a less obvious use of the product rule when taking the derivative of $f(x) = x^2$. The derivative is simple, if we use the power rule we get: $$\frac{d}{dx} f(x) = 2x$$

And this is done with the formula $\frac{d}{dx} [x^n] = nx^{n-1}$.

But I am positing that there is a "hidden" product rule being performed as well! Because if we think about it, there is still multiplication taking place with the $x$ in $f(x) = x^2$, this is shown in the "expanded" version of: $$f(x) = 1\times x^2$$ And you cannot just skip over the product rule! So in taking the derivative of this function we are actually performing a combination of the power rule and product rule: $$\frac{d}{dx} f(x) = 0\times x^2 + 1\times 2x$$

which is implicitly done with the output of the power rule alone, so this is why we do not show the product rule in action here.

This is my own intuition based on what I have learned so far, and I am curious if I am correct, or making dangerous assumptions!

Ted Shifrin
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ZenPyro
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    You seem to be confusing the chain rule with the rule for differentiating a product. None of your examples use the chain rule since none concern a composition of functions. – Ethan Bolker Nov 27 '23 at 20:27
  • @EthanBolker and Ted. Sorry! I was also looking at the chain rule and I accidently was using "chain rule" instead of "product rule"! I am making an edit right now! – ZenPyro Nov 27 '23 at 20:28
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    You don't need the product rule to differentiate $cf(x)$. The constant rule is more basic and quite immediate. However, if you have not already seen this, you should prove the power rule (the derivative of $x^n$) by using the product rule and mathematical induction. – Ted Shifrin Nov 27 '23 at 20:35
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    The statement "you cannot just skip over the product rule" leads to a problem, no derivative will be computable, since in applying the product rule to $1\times f(x)$ you need to compute $$1\times f'(x)$$, but if we "have to use the product rule" then to compute the $f'(x)$ part of that expression, we need to apply the product rule again to $1\times f(x)$, and so the process repeats – Carlyle Nov 27 '23 at 20:49
  • @Carlyle Hm, maybe I am misunderstanding your reasoning, but you would only have to apply the product rule once, right? Say we had a more conventional use of the product rule: $f(x) = x^2\times sin(x)$, well we wouldn't get $f(x) = 2x\times sin(x) + x^2\times sin(x)$, and then keep deriving deeper, that would just be taking the second derivative, right? You do not take the derivative of the resultants of the product rule. – ZenPyro Nov 27 '23 at 21:02
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    @ZenPyro what I am saying is that if we demand the product rule be used whenever we take a derivative, by applying it to $1\times f(x)$ where $f(x)$ is the function we are differentiating, then we run into a problem, since then $$ \frac{d}{dx}f(x)= f(x)\times \frac{d}{dx} 1 + 1\times \frac{d}{dx} f(x) $$ but now by our rule, we need to compute $$\frac{d}{dx} f(x)$$ by applying the product rule, so we get $$f(x)\times \frac{d}{dx}1 + ( f(x)\times \frac{d}{dx} 1 + 1\times \frac{d}{dx} f(x) )$$, so we never make progress, unless we skip the product rule in the cases where it is not necessary – Carlyle Nov 27 '23 at 21:08
  • @Carlyle Hmm, you bring up an interesting point! But the only problem I see with this is, it is akin to saying: "The derivative of a function $f(x) = x^3$ is taken by applying the power rule, which results in $f'(x) = 2x^2$, and since there is still a power for $x$, we must continue taking the power rule". I know it is not exactly as you say, but when I said "you cannot just skip over the product rule", I meant in respect to the fact that you must apply all rules when taking the derivative of a function. – ZenPyro Nov 27 '23 at 21:23
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    @ZenPyro that is not the same as what I am saying, that would be incorrect indeed, but the product rule differs from the power rule in that it reduces the derivative to a linear combination of different derivatives, so I am not differentiating twice, I am simply doing what the product rule tells me to do, it tells me I need to take the derivative of both factors, but my original problem was taking the derivative of the second factor, so if I insist on doing the product rule whenever I see a derivative, I never stop doing product rules – Carlyle Nov 27 '23 at 21:28
  • @Carlyle I like this point, if you think of anything else interesting down the line, please let me know! – ZenPyro Nov 27 '23 at 22:14
  • @Carlyle I did some thinking, and I believe that your calculations are naively founded, as in, if you applied your thinking (of every derivative needing to be broken down into its parts and recursively apply the rule) to another conventional differential rule such as the chain-rule it would turn into infinite regress as well. Let me explain with an example: – ZenPyro Nov 28 '23 at 04:52
  • @Carlyle If we have a composite function $$f(g(x))$$ then to take its derivative, we would need to utilize the chain-rule, this is done as so: $$\frac{d}{dx}[f(g(x))] \cdot \frac{d}{dx}[g(x)]$$ Now if we applied your incorrect logic of the recursive breakdown of the chain-rule, then we would have to do a recursive chain-rule: $$(\frac{d}{dx}[f(g(x))] \cdot \frac{d}{dx}[g(x)]) \cdot \frac{d}{dx}[g(x)]$$ And by this logic, this would continue in infinite regress, with no sign of when to stop. So why is it that you are okay with putting aside this logic for the more conventional chain-rule -> – ZenPyro Nov 28 '23 at 05:01
  • @Carlyle -> but not this less conventional "hidden" product rule. Also I want to say, I still like your point, even if I do not agree with it. It was an interesting thought experiment like my initial question! – ZenPyro Nov 28 '23 at 05:06
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    @ZenPyro yes you are correct that this is a problem for the other rules as well, that is precisely why I pointed it out, from your statement "we cannot just skip the product rule" it seemed you were insisting that we do need to do this whole computation, but in fact there are times where we should "skip" the product rule – Carlyle Nov 28 '23 at 07:44
  • @Carlyle I see your point! But also if we think about what derivatives are telling us, this makes sense, it is saying that the rate that function $f$ is changing with respect to $x$. Which really means, what is the rate of change $f$ in terms of $g$ and $g$'s rate of change in terms of $x$, now we can see that we get the rate of change in $f$ in terms of $x$ by the products of two rates of change. A good example I saw is: "If a car travels twice as fast as a bicycle and the bicycle is four times as fast as a walking man, then the car travels 2 × 4 = 8 times as fast as the man." – ZenPyro Nov 28 '23 at 22:56

2 Answers2

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Great question. Your assumptions are correct. Your work isn't too important for derivatives, but when you get to integral calculus keep your intuition in mind and you will be ahead of your class. I will put how in a spoiler below, but I highly suggest you don't click on it unless you're burning to find out. Also, your intuition is probably the way the person who first made the product rule tested whether it worked or not.

Integration by parts is the product rule for integrals. If you write $\int f(x)dx$ as $\int 1\times f(x)dx$ and repeatedly integrate $1$ and differentiate $f(x)$, you get Taylor's Theorem.

Kamal Saleh
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  • Thank you for the answer, that is great to know! I'm sorry, but curiosity got the better of me, time to go deeper down the rabbit whole into the Taylor's Theorem! – ZenPyro Nov 27 '23 at 20:41
  • Actually I have one more quick question, if you don't mind answering! How come for a function that is the sum of two "portions" $f(x) = x + sin(x)$, we call $x$ and $sin(x)$ terms but for a function that is the product of two "portions" $f(x) = x\times sin(x)$, we call $x$ and $sin(x)$ functions? I feel like it would make more inutive sense, to always call the variables either terms or functions, since the operator does not change what $x$ represents, $x$ could be a constant or a function, but the sign does not seem to play a role in deciding that. – ZenPyro Nov 27 '23 at 20:46
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Rather than saying there's an implicit product rule taking place in these derivations, I would say that it demonstrates that the product rule is consistent. In other words, we can derive a lot of these results without using the product rule, but if you do use the product rule you get the same answer. Which is good, because if you do something two ways and get different results it's usually a sign that something's not right.

However, in my opinion, the more fundamental rule in calculus is the chain rule. The reason I say that is because starting with the chain rule, the linearity of the derivative (i.e. $(af(x) + bg(x))' = af'(x) + bg'(x)$) and the derivatives of the exponential and constant functions, you can derive a multitude of other results including the product rule and the derivatives of polynomials and trigonometric functions.

ConMan
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  • I love this answer! Super good insight! This actually plays into another one of my questions, if you don't mind answering! How come for a function that is the sum of two "portions" $f(x)=x+sin(x)$, we call $x$ and $sin(x)$ terms but for a function that is the product of two "portions" $f(x)=x×sin(x)$, we call $x$ and $sin(x)$ functions? I feel like it would make more intuitive sense, to always call the variables either terms or functions, since the operator does not change what $x$ represents, $x$ could be a constant or a function, but the sign does not seem to play a role in deciding that. – ZenPyro Nov 27 '23 at 22:57
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    That's mostly just convention - we tend to use "terms" to represent separate objects being added together, as in "the terms of a sequence / series". That doesn't stop them from also being functions. We could also use "terms" when referring to a product, we just tend not to for whatever reason. – ConMan Nov 27 '23 at 23:13