Question

Differentiation anyone?

Answer 1

\[\LARGE y=(x+x^{-1})^3\]

Answer 2

Without chain rule.

Answer 3

expand the polynomial and use the quotient rule?

Answer 4

technically we are using the chain rule on y = x^2 , the y'=2x*1 = 2x

Answer 5

We're treating it as if we just learned the power rule and no other.. yea I know my fault for failing the first time -.-

Answer 6

But when expanding with negatives.. we still add them right?
\[\LARGE (x^a)(x^b)=x^{a+b}\]

Answer 7

yes

Answer 8

so we wont even need the quotient rule,
\((x+x^{-1})^2=x^2+x*x^{-1}+x*x^{-1}+x^{-1}*x^{-1}=x^2+2+x^{-2}\)

Answer 9

Use binomial theorem: \[
(a+b)^n=\sum_{k=0}^n{n\choose k}a^nb^{n-k}
\]

Answer 10

We can do pascal triangle:
1
1 1
1 2 1
1 3 3 1

Answer 11

yeah, there is a very fast way of using the binomial theorem, you can do (a+b)^10 just as fast as you can write it down.

Answer 12

even faster than having to think of the triangle. google the method and practice it and you can amaze your friends.:)

Answer 13

https://www.khanacademy.org/math/trigonometry/polynomial_and_rational/binomial_theorem/v/binomial-theorem--part-2
go to the 7:40 mark
its very cool

Answer 14

In this case we can say \[
(a+a^{-1})^n = \sum_{k=0}^{n}{n\choose k}a^n(a^{-1})^{n-k}=\sum_{k=0}^{n}{n\choose k}a^na^{k-n}=\sum_{k=0}^{n}{n\choose k}a^k
\]

Answer 15

So we got \[
1+3x+3x^2+x^3
\]Hmmm, this is hard to actually believe, lol...

Answer 16

Yes, what I put is wrong.

Answer 17

I was wondering what happened to the negatives xD
Thanks tho, I got it from here~

Answer 18

Correction: \[
(a+a^{-1})^n = \sum_{k=0}^{n}{n\choose k}a^k(a^{-1})^{n-k}=\sum_{k=0}^{n}{n\choose k}a^ka^{k-n}=\sum_{k=0}^{n}{n\choose k}a^{2k-n}
\]
Exponents are \(2(0)-3=-3\), \(2(1)-3=-1\), \(2(2)-3=1\), \(2(3)-3=3\)
We get: \[
x^{-3}+3x^{-1}+3x+x^{3}
\]

Answer 19

you should really check that video:) as fast as you can type it.....