A math "thought" question (not homework help)...

Pre-College Issues High School Life
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<p>This post isn't intended to be an inquiry for homework help. My teacher didn't actually ask me or anyone in my class to do or think about this for homework, but I remain rather perplexed and have sought the collective wisdom of CC to help me reconcile my confusion :). If this post is nevertheless interpreted or deemed as such then I guess it'll have to be deleted or moved, oh well...</p>

<p>So anyways, I recently got back an AP Physics quiz on basic derivatives. Most of the kids in my class are either in AB or BC Calc, and I took BC last year and am soon going to start self-studying a bit of MV Calc, so I have more experience with derivatives than the rest of the class and have gotten pretty good at them. On this quiz though I got docked a few points more than the rest of my peers for performing a substitution too soon, in the paraphrased words of my teacher. Here's what happened:</p>

<p>The quiz asked me to derive y=cos(ln(x)). Easy enough right? I substituted u for ln(x), rewrote the equation as y=cos(u), and said (dy/dx)=(dy/du)*(du/dx). All was well thus far...</p>

<p>I then said that (du/dx)=(1/x). Check. Then I said (dy/du)=-sin(u). Then came my supposed flaw for which I lost some credit. On the next line I re-substituted ln(x) for u, writing it: (dy/du)=-sin(ln(x)). Red circle -> "substitution happens later." I ended up, of course, with -sin(ln(x))/x as my final answer, but I did not receive full credit because of my apparent miscue.</p>

<p>Now I really don't care about a few trivial points here and there (truth be told, I've already resolved my college plans and so I can afford a little grade leeway, so to speak). But I didn't think I was necessarily wrong. So I approached my teacher and gave him a tedious and to-the-point argument: "(dy/du)=-sin(u). -sin(u)=-sin(ln(x)). (dy/du)=-sin(u)=-sin(ln(x)). Therefore (dy/du)=-sin(ln(x)). An equality is an equality. Why does it matter when I choose to ultimately substitute?" To this he said, "You can't do that because (dy/du) has to equal a function with u as its variable." Upon pressing and saying "but u IS ln(x) written in a different way," he said that he couldn't argue the point with me but that set theory and further uncovering the true meaning of the derivative in college math would show that I would be wrong." Now that was rather unsatisfying, but I suppose I was somewhat willing to take his word on it, just because. But that presented me with my current dilemma (yeah yeah, I only just got there. I am a little restless at this point. Bear with it!):</p>

<p>The way I see it, there are 3 possibilities:
1) He's wrong and I'm right. There is nothing mathematically incorrect about writing (dy/du) as equal to -sin(ln(x)). An equal sign is an equal sign, nothing more, nothing less.
2) I'm wrong and he's right. This has something to do with set theory, which is a topic that is rather beyond my head at this point. I'd love to hear an explanation of why set theory disproves my notion though, from the bright folks at CC, some of whom I'm sure have had the ambition to get that high up in math.
3) We're both wrong; or rather, we're all wrong in a sense. (dy/du)=-sin(u), -sin(u)=-sin(ln(x)), -sin(ln(x))=/=(dy/du). If the latter two are true, then (dy/du) isn't truly equal to -sin(u) in the sense that (dy/du) isn't a function and only a function like the other two are but rather a definition, a vocabulary term, etc. This is something hard for me to put into words but basically what I'm saying is that they aren't equal so much as **equivalents in different lingo<a href="-sin(u)%20is%20strictly%20function-speak,%20whereas%20(dy/du)%20is%20a%20more%20formalized%20term">/b</a>. Maybe something more appropriate would be "(dy/du) is defined as -sin(u)." Maybe that wouldn't be enough of a distinction from "equal." Maybe I'm just complete freaking nuts right now. Maybe you guys can effectively dismantle all of the thoughts I've presented in this post in a handful of words. </p>

<p>Whatever your contribution is, it would be greatly appreciated! Again, I hope it can be appreciated that this isn't a homework help problem. If there is a more appropriate forum in which I can post this thread though, I will gladly. I figured though that with all the high school kids in here tackling calculus this would be a good place to start. So yeah, thanks!</p>

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<p>Good question. I can’t answer it myself, being in AB, but I would like to hear what others have to say.</p>

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<p>I have to take care of something right now, but I will be back to reply.</p>

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<p>Woah, I’m in BC calc this year and have been doing this sort of stuff for all of this year and several months of last year, and that’s not how I’ve seen the chain rule at all. We’re just assumed to know to use it when there’s a function within a function and you can just write it without all of those intermediate steps…</p>

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<p>^ That’s a systematic way of looking at it.
If I wasn’t tired as all get out, and doing other things right now, I’d go look through my set theory notes and examine it, but this is a good thought.</p>

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<p>Yeah, I don’t like doing the chain rule that way myself. But my teacher pretty much requires us to go about it that way.</p>

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<p>^^I haven’t heard that before; does that just mean it’s doing it from memorization as opposed to figuring it out on the spot/proving it each time?</p>

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<p>That’s not what I intended to mean. I mean once you got it, you got it. But I don’t like just assuming. That’s not what calculus is about to me.</p>

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<p>lol, who uses substitution to do the chain rule?</p>

<p>Anyway, I’m pretty sure he’s wrong. this has nothing to do with set theory</p>

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<p>Yeah why use substitution to do the chain rule?</p>

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<p>^ Because my teacher is a stinkyhead! (Not really, he’s just really particular about things…I have a feeling he’s doing a lot of hand-holding for the kids who’ve never really dealt with any calc before until this year, and so it’s moving slowly right now)</p>

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<p>Since dy/du is the rate of change in y with respect to u, it would only make sense when you are using u as a variable. Still, I don’t think that there is anything wrong with your response. It is mathematically equivalent.</p>

<p>I assume that your teacher wanted you to perform the substitutions all at once:
(dy/du) = -sin(u); u = ln(x); du/dx = 1/x
(dy/du)(du/dx) = (-sin(ln(x)))(1/x)</p>

<p>However, I don’t think that method is necessarily more correct. I also don’t see the connection to set theory.</p>

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<p>

Differentiation of parametric equations?</p>

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<p>y = cos(ln(x))
y’ = - sin(ln(x)) * 1/x
y’ = -sin(ln(x))/x</p>

<p>You’re wrong for not realizing how easy this problem is. What were you doing trying to use substitution?</p>

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<p>^

</p>

<p>Also, your method is substitution in shorthand, I think.</p>

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<p>^^Yeah, you’d have gotten even more points docked off by that teacher than OP did :D</p>

<p>That seems to work…there’s no rule against substituting before you get to the end of the problem. f(u)=f(u(x)), right, because u=u(x). And 99 times out of 100 it makes it a lot easier to deal with too. The only time anyone ever advocated keeping the u in was when we didn’t know what u(x) was and you had to write the general form of the chain rule. </p>

<p>However, I do understand where he’s coming from on dy/du=sin(ln(x)), which is really not true. When you write d/du, you’re assuming u is independent. You sort of want to get it in terms of x, so dy/dx=sin(ln(x))(1/x) and dy/du…sort of…is just the sin(ln(x)) part of that, but in terms of u. I don’t know why he cared that you substituted after you had differentiated. But still, this is something I know only because a mathematician ranted at me about it. I don’t think this particularly matters in an AP Physics class.</p>

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<p>^ Frankly though you basically just gave the same argument my physics teacher gave. It doesn’t change the fact that mathematically they are equal…(unless someone has some set theory or other magic to work!). This mathematician, when he ranted at you about the same issue, how did he reconcile the fact that these terms were all equal to each other?</p>

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<p>Mathematically equal. The substitution doesn’t change the answer at all. Your physics teacher just has issues of style. dy/du as a function can still substitute a function of x in. Yeah, I think you’re right:
dy/dx=dy/du<em>du/dx
sin(ln(x))/x=dy/du</em>(1/x)
dy/du=sin(ln(x))</p>

<p>Now that I’ve actually thought about the question, rather than just rambling about it…</p>

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<p>what happened to the option that you’re both right?</p>

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<p>Mathematically that would be a contradiction, no? :p</p>