Does anyone have any nice ways of handling differentials like \[
\frac{d^2}{dx^2}\left(\frac{20}{1+x^2}\right)\]? And, the more extreme cases like \[\frac{d^4}{dx^4}\left(\frac{20}{1+x^2}\right)\].

- anonymous

- schrodinger

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- eSpeX

Have you considered using a negative exponent?
\[\frac{d^2}{dx^2}(20)(1+x^2)^{-1}\]

- anonymous

Yes, but the problem gets ugly, quickly, especially considering we are taking the second derivative. So, I'm guessing, there is none? All just straight up computation. Hmm.

- lgbasallote

i usually use a "u-substitution" in derivatives...i suppose that makes me weird....

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## More answers

- anonymous

Enlighten me...?

- calculusfunctions

@LolWolf \[\frac{ d ^{2} }{ dx ^{2} }(\frac{ 20 }{ 1+x ^{2} })\]implies that you find the second derivative of the function y, if\[y =\frac{ 20 }{ 1+x ^{2} }\]I assume you know your derivative rules, correct?

- anonymous

Yes, I'm not wondering so much how to find it, but, rather, how to do it neatly.

- anonymous

(If someone has some ingenious way of managing it, et al)

- anonymous

y'=-40x/(1+x^2)^2 now do it again the second time

- calculusfunctions

No, I am a math teacher and I will tell you that you must use the quotient rule to find the second derivative, but it would be wise to use the chain rule (power of a function rule) to find the first derivative.

- anonymous

use the quotient rule

- anonymous

@mark_o. Yes, I know. I'm not talking about how to find it, finding it is simple, I'm referring to taking this twice or thrice without having an utter mess of calculations, or without it taking more than some short time. I don't know if there even *is* such a way, I'm just guessing.

- calculusfunctions

@mark_o. please stop giving out answers. You're not helping!

- anonymous

I already have the answer. That's not what I'm looking for. I'm wondering if there is some way to manage this in a quick and nice manner, rather than crunching numbers.

- anonymous

(Again, there might *not* be, but I'm not the most imaginative person in Calculus--I deal with Number Theory--and this question came up)

- calculusfunctions

Those derivative rules are in place for a reason. You're looking for a short cut and I'm telling you that those derivative rules are the shortcut.

- anonymous

im sorry i didnt want to give the answer on the second derivative., i thought he didnt knoe how to arrive at the first derivative.... continue plzz

- calculusfunctions

@mark_o. That's alright! He said he already got the answer so no harm done.

- anonymous

Yes, I am not referring to the derivative rules. Of course, most people could come up with the proofs, and I am not looking for a shortcut, at least, not directly. The point that I ask is *not* how to find them, nor to get a lesson on the shortcuts, and whether they should be used or not, I'm asking if there *is* a nice way of *handling* the operations, or if there is some much nicer (not necessarily *easier*) way of dealing with problems that are similar.

- anonymous

ok continue plzz...

- calculusfunctions

@mark_o. there is nothing to continue LOL because @LolWolf the answer to your question is a very nice NO.

- anonymous

That's all I wanted to know. Thank you.

- calculusfunctions

Welcome!

- anonymous

well, there no way of getting around it,or to short cut them, just continue solving them wether you started on quotient rule or product rule.you will arrive on the same answer,......

- anonymous

Yes, thank you.

- anonymous

ok YW, have fun now

- anonymous

Actually, there IS, indeed a much nicer way to do this derivative, if we split it into separate functions. Not to beat a dead horse, or anything, but please don't say that one 'cannot' do something without actually knowing so, even if you're a teacher.
If we simply assume the function of \(x^2=l\), then, by repeated application of the chain rule, we reduce the laborious task of handling the repeated quotient rule to a simple product rule with a single multiplication per step, along with a much easier-to-compute chain rule.
Enjoy.

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