Expert: Paul Klarreich - 9/26/2009

Question
QUESTION: Hi.
I want to write a proof for the derivative of x^n as n*x^n-1 with the binomial theorem

my work:          

d/dx  a(x^n)

(1) a (lim dx >- 0 (x+dx)^n -(x)^n / dx

(2) a lim dx>-0 (nC0 x^n + nC1 x^(n-1)dx +nC2 x^(n-1)dx^2..... nCr dx^n  - (x)^n ) /dx

(3) a lim dx>-0   nC1 x^(n-1) dx + nC2x^(n-2)dx^2 ....nCr dx^n / dx

(4) a lim dx>-0  dx(nC1 x^(n-1) + nC2x^(n-2) dx ... nCrdx^n-1) /dx
(5) a lim dx>-0   (nC1 x^(n-1) + nC2x^(n-2) dx ... nCr dx^(n-1)
(6) a  (nC1 x^n-1) +0
(7) a (n!/1!(n-1)! x^n-1)
(8) a (n (n-1)!/(n-1)! x^n-1)

a n*x^n-1

Is this all correct
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(sorry If I maybe didn't follow your instructions I didn't understand (2) and for (7) I can't be exact as I am hometeaching myself algebra and calculus)

ANSWER: Questioner: Hamad
Country: Kuwait
Category: Calculus
Private: No
Subject: d/dx(x^n) proof with binomial theorem
Question: Hi.
I want to write a proof for the derivative of x^n as n*x^n-1 with the binomial theorem

my work:          

d/dx  a(x^n)

>> I think you can skip the a's

(1) (lim dx >- 0 (x+dx)^n -(x)^n / dx

(2) lim dx>-0 (nC0 x^n + nC1 x^(n-1)dx +nC2 x^(n-1)dx^2..... nCr dx^n  - (x)^n ) /dx

>> Small misprint there --
   your last term is  nCn dx^n, which is  1 dx^n

(3) a lim dx>-0   nC1 x^(n-1) dx + nC2x^(n-2)dx^2 ....nCn dx^n / dx

(4) a lim dx>-0  dx(nC1 x^(n-1) + nC2x^(n-2) dx ... nCdx^n-1) /dx

(5) a lim dx>-0   (nC1 x^(n-1) + nC2x^(n-2) dx ... nCn dx^(n-1)

(6) a  (nC1 x^n-1) + 0

(7) a (n!/1!(n-1)! x^n-1)

(8) a (n (n-1)!/(n-1)! x^n-1)

a n*x^n-1

Is this all correct.

>> Basically, yes.  You might look for ways to simplify some of the work.
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(sorry If I maybe didn't follow your instructions I didn't understand (2) and for (7) I can't be exact as I am hometeaching myself algebra and calculus)

>> You did fine.

NOW YOUR NEXT TASK WILL BE:

Use this format:
         f(r) - f(x)
f'(x) = lim[r -> x] -----------
         r - x

to prove this rule.  You will use your knowledge of factoring INSTEAD of the binomial theorem.  You should find it interesting and useful.


---------- FOLLOW-UP ----------

QUESTION: How can I factor r^n-x^n / r-x ?

ANSWER: QUESTION: How can I factor r^n-x^n / r-x ?

I thought you would get that -- perhaps you gave up too quickly, but:

Observation 1: The Factor theorem says that P(a) = 0, then (x - a) is a factor of P(x).

Application here:  (r - x) is always a factor of  r^n - x^n  

Observation 2: Check out these factorings:

r^2 - x^2 = (r - x)(r + x)
r^3 - x^3 = (r - x)(r^2 + rx + x^2)

Do r^4 - x^4 yourself, and deduce the pattern.  Generalize this and try to finish up.


---------- FOLLOW-UP ----------

QUESTION: Ok so this is what I tried to deduce
r-x= r-x = (r-x) * (r^(n-1)x^0*x^(n-1)r^0
r^2 - x^2 = (r - x)(r + x) = (r-x) * (r^(n-1)x^0+x^1 r ^0)
r^3 - x^3 = (r - x)(r^2 + rx + x^2)=  (r-x) * (r^(n-1)x^0 + r^n-2 x^1+r^(n-3)x^2
r^4 - x^4 =(r-x) (r^3 +xr^2+rx^2 +x^3) = (r-x) * (r^(n-1) x^0 + r^(n-2) x^1+r^(n-3)x^2+r^(n-4)x^3


which means for every r^n-x^n it = (r-x) (r^(n-1)x^0......r^n-nx(n^-1)
so I think that the solution for lim x>r  r^n-x^n/r-x
=(r^(n-1)x^0+ r^(n-2)x^1......r^(n-n)x^(n-1)
=r^(n-1)r^0 + r^(n-2)r...... r^0*r^(n-1)
=r^(n-1)((r^(n-2)r+r^(n-3)r^2.....r*r(n-2))

idk maybe I did it all wrong but atleast I tried  

Answer
QUESTION: Ok so this is what I tried to deduce
r-x= r-x = (r-x) * (r^(n-1)x^0*x^(n-1)r^0
r^2 - x^2 = (r - x)(r + x) = (r-x) * (r^(n-1)x^0+x^1 r ^0)
r^3 - x^3 = (r - x)(r^2 + rx + x^2)=  (r-x) * (r^(n-1)x^0 + r^n-2 x^1+r^(n-3)x^2
r^4 - x^4 =(r-x) (r^3 +xr^2+rx^2 +x^3) = (r-x) * (r^(n-1) x^0 + r^(n-2) x^1+r^(n-3)x^2+r^(n-4)x^3


which means for every r^n-x^n it = (r-x) (r^(n-1)x^0......r^n-nx(n^-1)
so I think that the solution for lim x>r  r^n-x^n/r-x
=(r^(n-1)x^0+ r^(n-2)x^1......r^(n-n)x^(n-1)
=r^(n-1)r^0 + r^(n-2)r...... r^0*r^(n-1)
=r^(n-1)((r^(n-2)r+r^(n-3)r^2.....r*r(n-2))

ok maybe I did it all wrong but atleast I tried

Actually you did pretty well; you just have to find ways to reduce the writing (typing). When you get to be my age (don't ask) you'll know how.

You have:

which means r^n-x^n  = (r-x) (r^(n-1) + r^(n-2)x +  + rx^(n-2) + x^(n-1)

Now divide by (r-x), to get:

r^n-x^n
------- = (r^(n-1) + r^(n-2)x +  + rx^(n-2) + x^(n-1)
r - x

Now observe that:
1) There is a term for each  x^k, where  k = 0,1,2 up to n-1, which amounts to exactly  n terms.
2) When  r -> x, each term becomes exactly  x^(n-1).

So the right side becomes  n x^(n-1), the formula we were looking for.