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 Derivative x^n (Math | Calculus | Derivatives | Table Of | x^n)

## 3 Proofs

 x^n = n x^(n-1)

Proof of x^n : from e^(n ln x)

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

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)     Q.E.D.

Proof of x^n : from the Integral

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

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)     QED

Proof of x^n : algebraically

Given: (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)! )

Solve:

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)     QED