Let f(x)=x^n

Prove f^1(x)=nx^n-1 by definition

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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.

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.

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)

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

ok i will send it to you in a while.

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

Proof of derivative by definition.docx Finally

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Thank you! :)

Glad I 've helped.

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