# Derivative of a rational function

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?

Thanks.

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Are you confused about the quotient rule for derivatives or the algebra used in simplifying it? – Brandon Carter Feb 9 '11 at 17:17
I'm familiar with the quotient rule, I simply couldn't understand some of the simplifications made. :P – user6839 Feb 9 '11 at 18:57
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. – Bill Dubuque Feb 9 '11 at 20:46

\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! – user6839 Feb 9 '11 at 18:54
@user6839: In step 5 and 6? Yes, I do. Thanks. – Arturo Magidin Feb 9 '11 at 18:57
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! – user106148 Nov 7 '13 at 14:44

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! – user6839 Feb 10 '11 at 10:48
@user6939: It's simply the product rule $\rm\ (ab)' = a'b + ab'\$ for $\rm\ a = f,\ b = g^{-n}\:.$ – Bill Dubuque Feb 10 '11 at 16:25
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. :P – user6839 Feb 10 '11 at 17:09
@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. – Bill Dubuque Feb 10 '11 at 17:21
@Bill: is "elementary" being used in any technical sense here? – Pete L. Clark Jul 4 '11 at 19:37

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! – user6839 Feb 9 '11 at 17:37