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

3 Proofs

(d/dx) x^n = n x^(n-1)

Proof of (d/dx) x^n : from (d/dx) e^(n ln x)

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

(d/dx) x^n = (d/dx) e^(n ln x)
= (d/du) e^u (d/dx) (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 (d/dx) x^n : from the Integral

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

(integral)x^(n-1) dx = x^n / n
(d/dx) x^n / n = (d/dx)(integral)x^(n-1) dx = x^(n-1)
1/n (d/dx) x^n = x^(n-1)
(d/dx) x^n = n x^(n-1)     QED

Proof of (d/dx) 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:

(d/dx) 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

  
 
  

 
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