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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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    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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    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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    @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