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Calculus-based Rage.

edited October 2008 in Everything Else
I figure a forum of people devoted to various geekeries have the best chance of sympathizing with my rage at the situation I am faced with.

Today in calculus class, we were discussing derivatives, and my teacher put a simple function on the board:

f(x) = ax +b

...and asked us to find the derivative of said function. I raised my hand for the first time in a long time (I am not so great at calculus, and had recently obtained some help from a friend in linear algebra), and explained what I'd learned: that according to the constant multiple rule and the rule of derivatives of constant fuctions, the derivative is "a." He paused for a moment and then joked that none of that was correct, before correcting himself and proceeding to obtain the same answer with the difference quotient. Prior to seeking some help from my friend, I found the difference quotient to be tedious, drawn-out, confusing, and not at all conducive to how I learn, with a pile of opportunities to make errors on top of that (depending, of course, on the complexity of the derivative). So, for the past 2 weeks, I have been doing all of my homework with the quotient-derived rules for differentiation. I found that calculus suddenly became much easier to understand and do, as well as noticed the inefficiency of finding derivatives using the difference quotient. As my friend said, "Its like using binomial expansion when you know Pascal's Triangle."

This is about the point at which the rage kicked in. My teacher turned to me and said, "[WindUpBird], for the purposes of tomorrow's quiz, you will not be allowed to use any of these rules to find derivatives on tomorrow's quiz."

Me: *blink* "Wut?"
Him: "You need to know how to derive these rules for them to be of any use to you."

Needless to say, I was pretty upset. Not only did half an hour of tutoring from a friend teach me more than 2 months in this man's class, but isn't it beside the point how I find said answer, provided I show proof of how I came to my answer? According to my book, said rules are BEYOND what the rest of the class should know currently, so haven't I in fact surpassed expectations by learning this other method and applying it? Finally, why should I be forced to use a method that I have found to be directly contrary to the manner in which I learn?

As far as the ability to derive the rules goes, I really see no point. I achieve the goal, so why trouble myself with a method in which more opportunities for mistakes present themselves? I feel the same way about this as I do about screwdrivers: there is no reason why one should need to learn how to forge a screwdriver before one can USE a screwdriver.

So: can anyone tell me if this makes sense or not? Am I just being spited by a man who is upset that his quiz will be a piece of cake for me; or is he right, and I should force myself to attempt to learn in a fashion that confuses me and destroys my grades?
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Comments

  • Do this:

    Find your answer first with your nice method, but put it on the bottom of the page. Then go up to the top of the page and do it his way. If the answers are the same, you're good. If not, find out where you went wrong in the top, and if you can't, just change some of the numbers so the end result is the same as your way. Your teacher won't be able to discern the difference if you do your work in a non coherent, winding manner.
  • edited October 2008
    Use your rule to find the second derivative of sin(2x3).
    Post edited by HungryJoe on
  • To a certain degree your teacher is right, learning to derive your rules is an important part in understanding the the idea behind math - en ever increasing logical structure where all formulas are deducted from a few basic axioms (I know, Gödel).
    If I understood what subject you're learning, I assume he wants you to use simple algebra to subtract two infinitely close numbers in order to find the derivative of the function.
    You, on the other hand, want to say something like " f(x)=x" => f'(x)=1".
    You're right, but the theory behind this is too important to be neglected.
  • Using rules you don't fully understand is just as likely to lead you to a mistake as using a tedious but straightforward process.

    If all you care about is learning what rules to apply with no desire to understand why these rules work or where they come from, you may not be suited for math.
  • I should probably clarify: Despite not seeing the point in knowing how to derive them, I actually DO know how to derive them; we are also only working on the first derivatives of elementary functions.

    I suppose my problem is more that I feel like I can do something in a more straightforward manner, but I'm being prevented from doing so because the rest of the class hasn't gotten that far yet. Granted, said rules probably might be useless for higher derivatives, but in any event I already have the background to learn that, I've just discovered it differently.

    I've already started deriving all of the rules I know so that I can do it with ease tomorrow; however, I still feel like I should be able to apply what I know.
  • The rules that provide shortcuts for taking derivatives are easier and faster, but less straightforward (in the sense of coming directly from the defintion) than taking the limit of the difference quotient.
    Also, even something that looks as simple as the power rule can be tricky. There are different ways to prove it when the exponent is a positive integer, negative integer, rational number, or a real number in general.

    If you would like to prove the power rule for the case you wish to apply it on the top of your test, your teacher would certainly be unreasonable to not give you credit for knowing the material. However, unless you have been given a long string of repetitve problems, this is unlikely to save you time.
  • Alright, so I more-or-less see where everyone's coming from and have come to the conclusion that I was wrong. Thanks for the help.
  • edited October 2008
    Use your rule to find the second derivative of sin(2x3).
    I'm pretty sure you can get the derivative of that without using the difference quotient. I suck at doing derivatives with trig functions, so I'm not even going to try, but I'm still pretty sure you can do it the short way.

    EDIT: fuck it, I'll try.
    f(x)= sin(2x3)
    f'(x)= [sin(6x2)][cos(2x3)]
    f"(x)= I'm not doing it right now, I get enough calc homework as it is. It's going to be long and drawn out, but I assure you that you don't need to touch any (delta)x nonsense in order to do it.

    If that's wrong (and it probably is, it doesn't seem right), then whatever. I said I suck at doing derivatives of trig functions, and we haven't touched derivatives of trig functions in about a month.

    Like...for the first 2-3 we were doing derivatives, my teacher wanted us to use the difference quotient, but after that he didn't want to see it again, because he agreed that it is confusing. For the quizzes where he wanted us to use the difference quotient, I just did this:
    Find your answer first with your nice method, but put it on the bottom of the page. Then go up to the top of the page and do it his way. If the answers are the same, you're good. If not, find out where you went wrong in the top, and if you can't, just change some of the numbers so the end result is the same as your way. Your teacher won't be able to discern the difference if you do your work in a non coherent, winding manner.
    Post edited by Dkong on
  • edited October 2008
    Use your rule to find the second derivative of sin(2x3).
    I'm pretty sure you can get the derivative of that without using the difference quotient. I suck at doing derivatives with trig functions, so I'm not even going to try, but I'm still pretty sure you can do it the short way.
    Show me, then. Use the method described by Mr. Bird.
    Post edited by HungryJoe on
  • edited October 2008
    Use your rule to find the second derivative of sin(2x3).
    I'm pretty sure you can get the derivative of that without using the difference quotient. I suck at doing derivatives with trig functions, so I'm not even going to try, but I'm still pretty sure you can do it the short way.
    Show me, then. Use the method described by Mr. Bird.
    Check my previous post, I just edited it.

    edit: I fucked up and realized my fuck up. Going to edit it now.
    Post edited by Dkong on
  • edited October 2008
    Use your rule to find the second derivative of sin(2x3).
    I'm pretty sure you can get the derivative of that without using the difference quotient. I suck at doing derivatives with trig functions, so I'm not even going to try, but I'm still pretty sure you can do it the short way.
    Show me, then. Use the method described by Mr. Bird.
    Check my previous post, I just edited it.

    edit: I fucked up and realized my fuck up. Going to edit it now.
    That's not what Mr. Bird described. You're trying to use the rule for finding the derivative of sin(u), but Mr. Bird was trying to describe the Power Rule. You screwed up your first attempt, and then couldn't figure out the second derivative because you don't know the Product Rule. Suddenly the difference quotient doesn't sound like such a bad idea, does it? Oh, but that must be because I'm old and all old people are stupid. Maybe it's because you youngsters are too lazy to use the difference quotient.
    f"(x)= I'm not doing it right now, I get enough calc homework as it is. It's going to be long and drawn out, but I assure you that you don't need to touch any (delta)x nonsense in order to do it.
    Translation: I can't do it.

    Oh yeah, that delta(x) nonsense? Try doing any real physics without it.
    Post edited by HungryJoe on
  • edited October 2008
    don't know the Product Rule
    f(x)=f(x)g(x)
    f'(x)=f'(x)g(x)+f(x)g'(x)
    STFU
    Translation: I can't do it.
    Translation, I SUCK BALLS AT DERIVATIVES OF TRIG FUNCTIONS. God damn.
    I'll do the problem properly in a few hours once I'm done with my real homework.
    Post edited by Dkong on
  • edited October 2008
    don't know the Product Rule
    f(x)=f(x)g(x)
    f'(x)=f'(x)g(x)+f(x)g'(x)
    STFU
    You obviously don't know what you're talking about because you need this, you see to calculate the second derivative of your first derivative, which was wrong, btw.
    Translation: I can't do it.
    Translation, I'm fucking sick of calculus. But since you're being a jackass and pushing me to do it, I'll probably end up whipping out my calc book tonight and finishing the problem.

    And seriously, I SUCK BALLS AT DERIVATIVES OF TRIG FUNCTIONS. Give me the hardest derivative you can that doesn't involve trig functions (or logs, too, I guess, since we never even covered those in the first place) and I'll do it. Otherwise you're gonna have to wait till I get my calc book out.
    If you can't do trig or logs, you must not know very much about Calculus.

    See, you little bastards don't seem to realize, either through laziness or stupidity, that the difference quotient is important because it can tell you an instantaneous rate of change. This will be important to you if you have any plans to understand university physics. If you approach physics thinking that the derivative is just a set of rules for making quick calculations, you'll never understand what's going on.

    Further, if you know how to use the difference quotient, you won't have to be worried about forgetting the hundreds of different rules you're going to see for obtaining derivatives. You'll be able to derive the rule from scratch.

    But NOOOOOOOO! You're a teenager! You obviously know more than your broken down old teacher. I mean he just has a college degree, and you haven't finished high school . . . Oh, wait.
    Post edited by HungryJoe on
  • edited October 2008
    You obviously don't know what you're talking about because you need this, you see to calculate the second derivative of your first derivative, which was wrong, btw.
    If that's wrong (and it probably is, it doesn't seem right), then whatever.
    If you can't do trig or logs, you must not know very much about Calculus.
    Eh, we're only in chapter 3. I'll get over it eventually. We never learned derivatives of logs, and we haven't touched derivatives of trig functions since chapter 1, so I've long forgotten about them.

    Anyways....
    f(x)= sin(2x3)
    f'(x)= [cos(2x3)][6x2] (I wasn't sure if it was chain rule or not...after consulting my book, guess it was the chain rule)
    f"(x)= [cos(2x3)][12x] + [-sin(2x3)][6x2][6x2]
    ......= 12x cos(2x3) - 36x4 sin(2x3)
    Post edited by Dkong on
  • edited October 2008
    Eh, we're only in chapter 3. I'll get over it eventually. We never learned derivatives of logs, and we haven't touched derivatives of trig functions since chapter 1, so I've long forgotten about them.
    Translation: I was shit talking when I said I could obtain the derivative in question because I don't know what I'm talking about.

    Anyways....
    f(x)= sin(2x3)
    f'(x)= [cos(2x3)][6x2] (I wasn't sure if it was chain rule or not...after consulting my book, guess it was the chain rule)
    f"(x)= [cos(2x3)][12x] + [-cos(2x3)][6x2][6x2]
    ......= 12x cos(2x3) - 36x4cos(2x3)

    amirite?
    -1 for not factoring the result.

    Also, that's not the chain rule.
    Post edited by HungryJoe on
  • edited October 2008
    f(x)= sin(2x3)
    f'(x)= [cos(2x3)][6x2] (I wasn't sure if it was chain rule or not...after consulting my book, guess it was the chain rule)
    f"(x)= [cos(2x3)][12x] + [-sin(2x3)][6x2][6x2]
    ......= 12x cos(2x3) - 36x4sin(2x3)
    Translation: suck my balls.
    Post edited by Dkong on
  • Translation: suck my balls.
    I don't think you want Hungryjoe to do that mate.
  • edited October 2008
    This thread has actually gotten me the post pissed off I've been at someone on a forum in over 2 years. Congrats, Joe.
    -1 for not factoring the result.
    Oh god...My excuse for that is that my teacher never made us do that. In b4 another "translation:" post. Save it. I don't give a fuck anymore.

    12x cos(2x3) - 36x4sin(2x3)
    12x [cos(2x3) - 3x3sin(2x3)]
    Post edited by Dkong on
  • edited October 2008
    This thread has actually gotten me the post pissed off I've been at someone on a forum in over 2 years. Congrats, Joe.
    I can't help it if you're a shit talker. That's a personal issue you have to work out for yourself.

    Also, you still haven't factored completely.
    Post edited by HungryJoe on
  • edited October 2008
    ....I'm stunned.
    I just did the fucking problem, which you said I couldn't do...and yet you still call me a shit talker?
    Also, that's not the chain rule.
    If I had a scanner, I'd scan the page in my book that says "Chain rule" at the top and then proceeds to give an example problem that is almost identical to the one you posted (except the one in the book didn't want a second derivative). I don't have a scanner, so I can't do that, though. You'll just have to take my word for it (and what does my word mean, I'm just a shit talker, right?)
    Also, you still haven't factored completely.
    12x [cos(2x3) - (3x3)sin(2x3)]
    Post edited by Dkong on
  • edited October 2008
    kong, you obviously couldn't do it at the time you said you could. Then you made a lot of excuses for not being able to do it. Then you obviously had some help from your book. Finally, you can't factor. That's not the same as doing it.
    Also, that's not the chain rule.
    If I had a scanner, I'd scan the page in my book that says "Chain rule" at the top and then proceeds to give an example problem that is almost identical to the one you posted (except the one in the book didn't want a second derivative). I don't have a scanner, so I can't do that, though. You'll just have to take my word for it (and what does my word mean, I'm just a shit talker, right?)
    Let me help you. This is the Chain Rule. This is the Product Rule. Can you see where you used the Product Rule instead of the Chain Rule?
    Post edited by HungryJoe on
  • edited October 2008
    kong, you obviously couldn't do it at the time you said you could. Then you made a lot of excuses for not being able to do it. Then you obviously had some help from your book. Finally, you can't factor. That's not the same as doing it.
    I actually did factor right. I just copy/pasted the part where I forgot to change cosine to sine when I was taking the second derivative. The final correct answer is right above your last post. If, by some reason unknown to me, there's more to factor...tough. That's is far as I'm taking it. You said it couldn't be done using "Mr. Bird's" method. I just proved that you can. Yes, it was messy, and yes I had to look at my book for some help, but the problem got solved without the difference quotient, which was the challenge in the first place.

    And finally, if you take a careful look at my first post in this thread. I never actually said that I could do it. I just simply said that I knew it could be done. The fact that I JUST did it is icing on the cake for me.
    Post edited by Dkong on
  • I actually did factor right. I just copy/pasted the part where I forgot to change cosine to sine when I was taking the second derivative. The final correct answer is right above your last post.
    Translation: After you told me I was wrong, I changed my answer and lied about not being finished.
    And again, my teacher has never made me factor completely.
    . . . but I thought all you kids were so much smarter than your teachers . . .
  • edited October 2008
    I'm not saying I'm smarter than you. I'm not saying I'm smarter than my teachers (though I honestly believe that I'm smarter than SOME of them, certainly not all, though). I'm saying that I finished your challenge (no matter how messy it was) and I'm happy with it.
    Post edited by Dkong on
  • edited October 2008
    kong, you obviously couldn't do it at the time you said you could. Then you made a lot of excuses for not being able to do it. Then you obviously had some help from your book. Finally, you can't factor. That's not the same as doing it.
    I actually did factor right. I just copy/pasted the part where I forgot to change cosine to sine when I was taking the second derivative. The final correct answer is right above your last post. If, by some reason unknown to me, there's more to factor...tough. That's is far as I'm taking it. You said it couldn't be done using "Mr. Bird's" method. I just proved that you can. Yes, it was messy, and yes I had to look at my book for some help, but by god I did the problem.

    And finally, if you take a careful look at my first post in this thread. I never actually said that I could do it. I just simply said that I knew it could be done. The fact that I JUST did it is icing on the cake for me.
    Kid, you didn't do what Mr. Bird was talking about. You looked in your book and found another rule. A rule that is completely different than what Mr. Bird ws trying to enunciate.
    The fact that I JUST did it is icing on the cake for me.
    After you went back and edited about five different times after I told you that you were wrong. That's some bitter icing.

    Your pride in this is just an indicator of how little you know.
    Yes, it was messy, and yes I had to look at my book for some help, but the problem got solved without the difference quotient, which was the challenge in the first place.
    No, that wasn't the challenge. The challenge was to do it with the Power Rule, which is what Mr. Bird was trying to use.

    Do you think anyone wants to use the difference quotient all the time? Of course there are simpe rules that derive from the difference quotient, but you need to understand the difference quotient first. You obviously don't.
    I'm not saying I'm smarter than you. I'm not saying I'm smarter than my teachers (though I honestly believe that I'm smarter than SOME of them, certainly not all, though).
    That's an indicator of maturity. Yes, I'm sure you're smarter than a lot of your teachers. That's why they have college degrees and you don't.
    Post edited by HungryJoe on
  • edited October 2008
    Yes, I'm sure you're smarter than a lot of your teachers.
    I love how you've been constantly putting words in my mouth this whole time. Note how I said "some" in my post, then you quote and say "a lot".
    Also, saying that a college degree = intelligence is about as naive as saying that having a high gpa = intelligence.
    After you went back and edited about five different times after I told you that you were wrong.
    When I did it the first time before looking at my book, I knew I was wrong (like when I had it with all the addition signs and shit). I even said, right after doing the problem, that it was wrong. You didn't tell me shit. When I noticed that I put cosine instead of sine at the end of the problem, I caught that mistake by myself. As for not factoring, I really don't give a shit. Factoring != completeness. That's like saying 2x+2y is less correct than 2(x+y). It's not. It's just a different way of writing that expression. You never specified if it needed to be factored until well into the "challenge". You also just said "use your rule to find the derivative of sin(2x3)". You never said "use your rule blah blah and get the answer in one try and factor it completely". Yes, it matters.
    No, that wasn't the challenge. The challenge was to do it with the Power Rule, which is what Mr. Bird was trying to use.
    He never said Power Rule. He said "derivative of a constant function" and "constant multiple rule".
    Post edited by Dkong on
  • edited October 2008
    Factoring != completeness. That's like saying 2x+2y is less correct than 2(x+y).
    You're the naive one if you think this doesn't matter. That's like saying spelling doesn't matter. Try integrating with trig identities if you can't factor. Try integrating using the partial fraction method if you can't factor. Try integrating the expressions you'll encounter in Differential Equations if you can't factor. You won't be able to, but that's okay. It's doubtful that you'll ever get that far.
    When I did it the first time before looking at my book, I knew I was wrong (like when I had it with all the addition signs and shit). I even said, right after doing the problem, that it was wrong. You didn't tell me shit. When I noticed that I put cosine instead of sine at the end of the problem, I caught that mistake by myself.
    It's odd that you went back and edited your posts right after I told you that you were wrong then. I guess it was just coincidence, huh?
    No, that wasn't the challenge. The challenge was to do it with the Power Rule, which is what Mr. Bird was trying to use.
    He never said Power Rule. He said "derivative of a constant function" and "constant multiple rule".
    If you don't understand from his post that he was talking about the Power Rule, then you don't understand much. That's not a surprise, since you don't know the difference between the Chain Rule and the Product Rule. You're probably not going to get a good grade in Calculus. Oh, but I guess that's okay, because you'll be one of those sour grapes types who'll say that grades != intelligence. (I'll let you in on a secret. That's what stupid people say to themselves because they can't make good grades.)
    Post edited by HungryJoe on
  • Try integrating using the partial fraction method if you can't factor.
    QFT. Solving recurrence relations by using generating functions is fun.
  • edited October 2008
    Let me help you.Thisis the Chain Rule.Thisis the Product Rule. Can you see where you used the Product Rule instead of the Chain Rule?
    I used the chain rule when deriving the first derivative. I then used a combination of the chain and product rule when deriving the second derivative.
    You're the naive one if you think this doesn't matter. Try integrating with trig identities if you can't factor. Try integrating using the partial fraction method if you can't factor. Try integrating the expressions you'll encounter in Differential Equations if you can't factor. You won't be able to, but that's okay. It's doubtful that you'll ever get that far.
    I'll wholeheartedly give you this one. I wasn't thinking of differentiation when I made my post. I was thinking of final answers only.
    It's odd that you went back and edited your posts right after I told you that you were wrong then. I guess it was just coincidence, huh?
    Believe it or not (in your case, you won't), I was constantly editing my posts without looking for your replies (until I felt satisfied with the current post I was editing). All of the posts I made in this thread where I was working on the equations were edited roughly 10-20 times each. There was one point where I left my equations with cosine instead of sine, and the only thing you said was wrong about it was "-1 for not factoring". You didn't even point out that I goofed and wrote cos instead of sin (in b4 "you just didn't know the correct derivative!").
    You're probably not going to get a good grade in Calculus.
    image
    I go to a private school, to get an A you need to be getting between 95-97%. That grade was from a few weeks ago when midquarter grades came out. Currently I'm getting a 92, which is a B+ at my school, would be an A if I was going to public school.
    If I'm really as dumb as you say I am, then grades certainly don't = intelligence.
    As for the next 3 quarters of Calculus, we'll see.
    Post edited by Dkong on
  • edited October 2008
    Sorry for being so harsh yesterday, it's not by any means wrong to want to use a simpler method to find an answer. Not wanting to use the limit defintion all the time is exactly why mathematicians have developed all these rules for how to find derivatives.
    However, trying to use them without proving them first, as many high school students tend to do, turns calculus into mostly pattern recognition. This may make things easier in the short term, but means that you will go into university-level classes lacking some important tools, especially if you plan on using the AP exam to avoid retaking the class.

    An example of a mistake you might make if you try to avoid difference quotients is trying to find the limit as x goes to 0 of (sin x)/x. If you did not use the sum angle formula and the difference quotient to take the derivative of sin x, you may be tempted to use L'hopital's rule without realizing the circularity of what you have just done.
    Post edited by csrjjsmp on
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