Proof of Derivative f(x)=x^b & f^1(x)=bx^{b-1}

[Oerg]

Hi,

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

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

 

[nicksauce]

It's easy if you expand (x+h)^n using the binomial theorem.
http://en.wikipedia.org/wiki/Binomial_theorem

See if you can get it yourself.

 

[Oerg]

erm the latex input isn't quite working out right in my first post sry

 

[mathboy]

It's easy if you expand (x+h)^n using the binomial theorem.
http://en.wikipedia.org/wiki/Binomial_theorem

See if you can get it yourself.

Doesn't that proof only work if the exponent is an integer?

 

[nicksauce]

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

 

[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?

 

[mathboy]

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

 

[Hurkyl]

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

 

[mathboy]

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

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

 

[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.