Prove that d/dx xⁿ = nxⁿ‾¹?

Question

Prove that d/dx xⁿ = nxⁿ‾¹, for all values of n.

^_^

Answer 1

Rewrite x^n=e^(n*ln(x))

d(x^n)/dx=d(e^(n*ln(x))) use e^u and u =n*ln(x) =>
d(x^n)=(n/x)*e^(n*ln(x))=
(n/x)*x^n=n*x^(n-1) qed.

Answer 2

Let f(x) = x^n (n is a positive integer)

Then f(x + h) = (x + h)^n
= x^n + n*x^(n-1)*h + n(n-1)/2!*x^(n-2)h^2 + n(n-1)(n-2)/3!3*x^(n-3)h^3 + ... + n*xh^(n-1) + h^n

So f(x + h) - f(x) = n*x^(n-1)*h + n(n-1)/2!*x^(n-2)h^2 + n(n-1)(n-2)/3!3*x^(n-3)h^3 + ... + n*xh^(n-1) + h^n

Whence {f(x + h) - f(x)}/h = n*x^(n-1) + n(n-1)/2!*x^(n-2)h + n(n-1)(n-2)/3!3*x^(n-3)h^2 + ... + n*xh^(n-2) + h^(n-1)

Mow f'(x) = lim h→0 [{f(x + h) - f(x)}/h] = n*x^(n-1) + 0 + ) + ) + .... + 0 + 0
= n*x^(n-1)

Now when n is negative say n = -p where p is a positive integer

f(x) = x^n = 1/x^p = u/v
f'(x) = (vu' - uv')/v^2
= -px^(p-1)/x^(2p)
= -px^(-p-1)
= nx^(n-1)

Suppose n is rational Say n = p/q where p, q are integers and q ≠0

Let y = f(x) = x^n = x^(p/q)
So y^q = x^p
so qy^(q-1) = px^(p-1)
So y^(q-1)y' = (p/q) x^(p-1)
So y^q/y * y' = (p/q) x^(p-1)
ie y' = (p/q) x^(p-1)*y/y^q
= (p/q) x^(p-1)*x^(p/q)/x^p
= (p/q) x^(p-1-p+p/q)
= (p/q) x^(p/q-1)
= nx^(n-1)

I do not know how to prove it for irrational n but above proves it is true for all rational n