How to prove the power rule (Calculus)?

Question

f(x) = x^n
f'(x) = nx^(n-1)

Anyone know how to prove this?

Answer 1

There are two approaches, depending on your background. If you're comfortable with the binomial theorem, do the following. Directly write out the derivative of f as a limit as h->0 of (x+h)^n-x^n divided by h. Use the binomial theorem to expand the (x+h)^n term. There will be an x^n term which cancels in the subtraction. There will be an nhx^{n-1} term, and there will be other terms with higher powers of h. When you divide by h, this yields nx^{n-1} and a polynomial divisible by h. When you take the limit as h->0, that polynomial goes away and you're left wtih nx^{n-1}.

If you're comfortable with induction, proceed as follows. You should induct on n. For n=1, f is a line, and we know it's slope is 1. Assuming inductively that the rule is true for k, then by the product rule, the derivative of x^{k+1}=x*x^k is equal to x^k+x*[deriv of x^k]. Since we assume the rule is true for k, we get x^k+x*[kx^{k-1}]. Do some algebra and you get (k+1)x^k. So it is true for k+1. By induction, we know it's true for all n.

Answer 2

If you want a first principles proof then you must use one of the definitions of differential which is
Limit(h approaches 0) of (f(x + h) - f(x))/h
Thus you need to work on ((x + h)^n - x^n)/h
You will need to expand the (x + h)^n using the binomial expansion so I assume that you must already have been taught this.

Answer 3

[(f(x+h) - f(x)]/h = [(x + h)^n - x^n]/h =
= (x^n + nhx^(n - 1) + .... + h^n - x^n)/h =
= nx^(n+1) + n(n-1)hx^(n-2) + ... + h^(n - 1)

Now, every term having an 'h' in it approaches zero as h approaches zero.