How can you prove f^1(x)=nx^n-1 by definition?

Calculus
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Let f(x)=x^n

Prove f^1(x)=nx^n-1  by definition 

Oct 30th, 2015

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Proof of.docx 

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Proof of   x^n 

1.Algebraically

If: (a+b)^n = (n, 0) a^n b^0 + (n, 1) a^(n-1) b^1 + (n, 2) a^(n-2)b^2 + .. + (n, n) a^0 b^n

Here (n,k) is the binary coefficient = n! / [k! (n-k)! ]

Solution:

  x^n = lim(d->0) ((x+d)^n - x^n)/d

= lim [ x^n + (n, 1) x^(n-1) d + (n, 2) x^(n-2) d^2 + .. + x^0 d^n- x^n ] / d

= lim [ (n,1) x^(n-1) d + (n, 2) x^(n-2) d^2 + .. + x^0 d^n ] / d

= lim (n,1) x^(n-1) + (n, 2) x^(n-2) d + (n, 3) x^(n-3) d^2 + .. + x^0 d^n

= lim (n, 1) x^(n-1) (all terms on right cancel out because of the d factor)

= lim (n, 1) x^(n-1) = n! /[1! (n-1)! ] x^(n-1) = n x^(n-1) 

2.from the Integral

If:  x^n dx = x^(n+1)/(n+1) + c;     Fundamental Theorem of Calculus.

Solution:

 x^(n-1) dx = x^n / n

 x^(n-1) dx = x^n / n

  x^n / n =   x^(n-1) dx = x^(n-1)

1/n   x^n = x^(n-1)

  x^n = n x^(n-1) 

3.From   e^(n ln x)

If:   e^x = e^x;   ln(x) = 1/x; Chain Rule.

Solution:

  x^n =   e^(n ln x)

=   e^u   (n ln x) (Set u = n ln x)

= [e^(n ln x)] [n/x] = x^n n/x = n x^(n-1)


Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 30th, 2015

by definition, i ment using f^1(x)= lim h->0   f(x+h)- f(x)/h

Oct 30th, 2015

ok i will send it to you in a while.

Oct 30th, 2015

I am trying to upload the file , but the system doesn't take it. I will keep on trying.



Oct 30th, 2015
Oct 30th, 2015

If you need anything else, please let me know.

Oct 30th, 2015

Thank you! :)

Oct 30th, 2015

Glad I 've helped.

Oct 30th, 2015

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