Proof of derivative

1. Dec 31, 2007

Oerg

Hi,

I was wondering if anyone could provide a proof for the following derivative

$$f(x)=x^b$$
$$f^1(x)=bx^{b-1}$$

Last edited: Dec 31, 2007
2. Dec 31, 2007

nicksauce

3. Dec 31, 2007

Oerg

erm the latex input isnt quite working out right in my first post sry

4. Dec 31, 2007

mathboy

5. Dec 31, 2007

nicksauce

One can use Newton's generalized binomial theorem, which is covered on the wikipedia page, for that case.

6. Dec 31, 2007

chickendude

Alternatively, you can use a couple of cases

Once you prove it for nonnegative integers, you can do this

For negative integers:
let $$n$$ be a positive integer

$$x^{-n} = \frac{1}{x^n}$$

taking the derivative of x^-n is the same as taking the derivative on the right via the quotient rule

$$\frac{d}{dx}\frac{1}{x^n}\right) = \frac{x^n(0) - 1(nx^{n-1})}{x^{2n}} = -nx^{-n-1}$$

proving it

Now you can prove it for rationals through this:

$$\frac{d}{dx} x^{p/q}$$

let $$y = x^{p/q}$$

$$y^q = x^p$$

implicitly differentiating

$$qy^{q-1} \frac{dy}{dx} = px^{p-1}$$

$$\frac{dy}{dx} = \frac{px^{p-1}}{qy^{q-1}} = \frac{px^{p-1}}{qx^{p(q-1)/q}}$$

$$= \frac{p}{q} x^{p/q - 1}$$

can you figure out a similar method for irrationals too?

7. Jan 1, 2008

mathboy

I don't think there's a way to prove it for irrational exponents using the binomial theorem. Something else must be used.

8. Jan 1, 2008

Hurkyl

Staff Emeritus
But once you know it for rational exponents, the proof for irrational exponents is typical of arguments in anaysis.

9. Jan 1, 2008

mathboy

That is the "something else". You cannot prove it directly with the binomial theorem.

10. Jan 1, 2008

Gib Z

Well to have used the binomial theorem to prove for rationals, you must have used the generalized version anyway, which applies to irrational exponents as well. You can prove it for rational exponents even without the binomial theorem, and after this its a simple matter of taking some limits.