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I've found the following derivative in my Calculus book and I can't get my my head around the algebra involved. Can anybody help me?

problem & solution

Thanks.

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    Are you confused about the quotient rule for derivatives or the algebra used in simplifying it?2011-02-09
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    I'm familiar with the quotient rule, I simply couldn't understand some of the simplifications made. :P2011-02-09
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    The only nontrivial simplification is factoring $\ 2x\:(x^2+1)\ $ out of both terms in the numerators of the 2nd term. You can avoid this by doing it generically first, i.e. instead of using the formula for the derivative of $\rm\ f/g\ $ instead work out the formula for $\rm\ f/g^n\ $, simplify it, *then* specialize $\rm\ f,g\:$. The simplifications are much easier *before* specialization.2011-02-09

3 Answers 3

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\begin{align*} \frac{d}{dx}\left(-\frac{x^2+1}{(x^2-1)^2}\right) &= -\frac{d}{dx}\left(\frac{x^2+1}{(x^2-1)^2} \right)&\quad&\mbox{(1)}\\ &= -\left(\frac{(x^2-1)^2(x^2+1)' - (x^2+1)\left((x^2-1)^2\right)'}{\left((x^2-1)^2\right)^2}\right)&&\mbox{(2)}\\ &= - \frac{(x^2-1)^2(2x) - (x^2+1)\left(2(x^2-1)(x^2-1)'\right)}{(x^2-1)^4}&&\mbox{(3)}\\ &= -\frac{2x(x^2-1)^2 - (x^2+1)(2(x^2-1)2x)}{(x^2-1)^4}&&\mbox{(4)}\\ &= -\frac{2x(x^2-1)^2 - 4x(x^2+1)(x^2-1)}{(x^2-1)^4}&&\mbox{(5)}\\ &= - \frac{2x(x^2-1)\left((x^2-1) - 2(x^2+1)\right)}{(x^2-1)^4}&&\mbox{(6)}\\ &= - \frac{2x(x^2-1)\left(x^2-1-2x^2-2\right)}{(x^2-1)^4}&&\mbox{(7)}\\ &= - \frac{2x(x^2-1)(-x^2-3)}{(x^2-1)^4}&&\mbox{(8)}\\ &= - \frac{2x(-x^2-3)}{(x^2-1)^3}&&\mbox{(9)}\\ &= -\frac{-2x(x^2+3)}{(x^2-1)^3}&&\mbox{(10)}\\ &= -(-2)\frac{x(x^2+3)}{(x^2-1)^3}&&\mbox{(11)}\\ &= 2\frac{x(x^2+3)}{(x^2-1)^3}.&&\mbox{(12)} \end{align*}

Notes.

  1. Pull out the minus sign fromt he derivative.
  2. Use the Quotient Rule.
  3. Do the derivatives in the numerator, using the Chain Rule for $(x^2-1)^2$.
  4. Finish the derivative.
  5. Do some of the algebra in the numerator. Notice that both summands in the numerator have a factor of $2x(x^2-1)$.
  6. Factor out $2x(x^2-1)$ from both summands in the numerator.
  7. Do the operations in the other factor.
  8. Do the algebra in the numerator.
  9. Cancel the $x^2-1$ in the numerator with one in the denominator.
  10. Pull out the minus sign from $(-x^2-3)$.
  11. Pull out the $-2$ from the fraction.
  12. Simplify $-(-2)$ to $2$, and rejoice for your answer matches the one in the book.
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    Wow, thank you! Don't you mean 2x(x² - 1) though? Anyway, that's exactly the algebra part I didn't get: I didn't notice I could factor that out. I did manage to solve the exercise by factoring (x² - 1) and dividing by the numerator though!2011-02-09
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    @user6839: In step 5 and 6? Yes, I do. Thanks.2011-02-09
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    the step 5 through six there is a rearrangement but it seems like 4x(x^2+1) is turned into 2(x^2+1) all of a sudden, i am a bit lost there could you please explain thank you!2013-11-07
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The only nontrivial simplification employed in your derivation is reducing the fraction by cancelling the common factor $\rm\ (x^2+1)\:.\:$ You can simplify this by first computing the derivative generically, i.e. compute the general formula for the derivative of $\rm\ f/g^n\ $, then perform the cancellation in the simpler general form, before specializing $\rm\:f,g\:$ to their values. Namely

$$\rm\displaystyle \bigg(\frac{f}{g^n}\bigg)'\ =\ \frac{f\:\:'g^n-n\:f\:g^{n-1}\:g'}{g^{2\:n}} =\ \frac{f\:\:'g-n\ f\ g'}{g^{n+1}}$$

Now, specializing $\rm\ n = 2,\ f = -x^2-1,\ g =\: x^2-1 \:,\: $ we find that the arithmetic is a bit simpler, since we have already cancelled the common factor $\rm\:g^{n-1}\:,\:$ it being glaringly obvious in the simpler generic form.

While this "generic preprocessing" is a bit trivial here, it can provide immense simplifications in other contexts,$\ $ e.g. $\: $ see this proof of $\ $ Sylvester's identity $\rm\ \ det(1+AB) = det(1+BA)\ $ that proceeds by taking $\rm\ det\ $ of $\rm\ (1+A\ B)\ A\ =\ A\ (1+B\ A)\ $ then generically cancelling $\rm\ det(A)\:.\ \ $ This cancellation of the "apparent singularities" where $\rm\:det(A) = 0\:$ is much less trivial than the cancellation of $\rm\: g^{n-1}\: $ in the above derivative calculation. Indeed, most non-generic proofs of Sylvester's identity usually resort to far less elementary non-algebraic methods to deal with such singularities (for example, topological proofs that appeal to ideas based upon density arguments). $\ $ Moral: a little generic thought can go a long way towards avoiding dense proofs.

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    I'm sorry, but I don't understand how you worked out that formula there. Would you care to explain? Thanks!2011-02-10
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    @user6939: It's simply the product rule $\rm\ (ab)' = a'b + ab'\ $ for $\rm\ a = f,\ b = g^{-n}\:.$2011-02-10
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    Wow, now I feel really stupid for not seeing that. That subtraction threw me off and I just crossed the product rule out right away. Thanks again for putting up with me. :P2011-02-10
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    @user6839: Don't sweat it - we all miss simple things now and then. When problem solving it is very important to learn to think at the "meta-level" as above. You'll learn much more this way than simply mimicing brute-force computations given as solutions in poorly written textbooks. While the difference may seem minor for simple problems like this it may prove crucial to success in less trivial problems, e.g. said purely algebraic proof of Sylvester's identity. Thus the easiest (mechanical) solution is not always the best pedagogically. Ditto for the "highest rated" solution.2011-02-10
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    @Bill: is "elementary" being used in any technical sense here?2011-07-04
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    @Pete In many senses, pedagogically, effectively (algorithmically), etc.2011-07-04
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    @Bill: Okay. So am I understanding properly: when you say "far less elementary", you are making a statement about your experiences in teaching and computer algebra and not a mathematical claim *per se*?2011-07-04
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    @Pete No. But based on earlier discussions it wouldn't suprise me in the least if your definition of "mathematics" is less encompassing than mine. E.g. as I recall, you've stated that you have little knowledge of effective mathematics - even in your own specialty (number theory). That's in stark contrast to my experience (and interests).2011-07-04
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    @Bill: So far as I am aware, I don't have a definition of mathematics. Is it necessary to do so in order to understand what you mean here? I am afraid I still don't get it. Are you stating a theorem of some sort? By the way, if something that I said before gave you the idea that I was uninterested in effectivity issues, I sincerely apologize. I have a research paper which is concerned with issues of algorithmic effectivity in arithmetic geometry. I would certainly be interested to hear explanations in this direction.2011-07-04
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    @Pete: If you don't have a definition of mathematics then how can it possibly make sense for you to ask if something is a "mathematical claim"? In any case, it's far too painstaking to attempt extended discussions in comments. Please feel free to either email me at my first.last@gmail.com, or to present your views in your own answer. PS I'm happy to hear that you've finally become interested in effectivity issues. Links?2011-07-04
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    @Bill: I'm sorry if I was unclear. I mean I do not have a *prescriptive* definition of mathematics. When I asked if you were making a mathematical claim, I don't mean anything philosophical here: I meant to ask if you are alluding to a theorem. For instance, if I were to say "Weil-Chatelet groups of abelian varieties over number fields are a deep and interesting field of mathematics" then I am not making a mathematical claim, whereas if I say "The Weil-Chatelet group of this rational elliptic curve has order $9$" then I am. I am sorry that I am doing such a bad job explaining myself.2011-07-04
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    @Pete Did you miss my prior remark about taking this meandering tangential discussion to email? Comments are not supposed to be used for such - as I'm sure you are aware.2011-07-04
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    @Bill: e.g. http://math.uga.edu/~pete/HasseBjornODD.pdf. My comments are devoted to trying to understand your answer. I'm sorry that you're not more interested in being understood: I will confess to suspecting that if I don't understand what you mean, some others won't either. Anyway, I gave this post a +1: except for that one phrase, the mathematics was clear and correct. Why you will not just say that your preference for one proof technique over another is a preference rather than a mathematical fact continues to puzzle me. I hope this will be toned down in future posts.2011-07-04
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    The irony is that I think the ideas / techniques you're propounding are fantastic. I just happen to like other arguments too and dislike telling others that they are wrong if they do not agree that $X$ is "much less elementary" than $Y$ or $Z$ is "dense", etc. I don't expect you to agree now -- you will have the last comment and I'm waiting to hear "nonsense", "ridiculous", etc. (That's not respectful discourse, by the way. But people quickly get used to you saying it.) But when you think further about it I hope you will see that you can easily present your answers so as not to do this.2011-07-04
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    @Pete Non sequitur. Suggesting to take this to email (per site rules) implies nothing about not being (more) interested in being understood. And, NO, I will most certainly *not* "tone down" my posts, since I strongly believe that such remarks are *absolutely essential* to teaching mathematics. Indeed, not too infrequently I receive emails from students thanking me for sharing such insights (even years down the road). If you disagree then you are welcome to post your own views. *Please* stop trying to censor mine. Finally, again, *please* stop attempting to put pejorative words into my mouth.2011-07-04
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Just try using the quotient rule for derivatives and see where you get stuck on the algebra. I'm guessing the book just skipped over some intermediate steps and that's why it's not immediately clear how it got to one step to the other. I suggest trying it out on your own first with pencil and paper, then ask again if you get stuck.

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    Yeah, it's exactly as you said. I managed to do the exercise on my own in a simpler fashion. Thanks!2011-02-09