# Derivative proof with a fractional exponent

1. Aug 21, 2006

### Chemical_Penguin

Hi all, I've been trying to get back into mathematics by teaching myself calculus. I've been starting with the book "Calculus Made Easy" and have been doing fine except for one little thing I encountered on page 57.

He shows a mathematical proof of why the derivative of

$$y = x^\frac {1} {2}$$

is...

$$\frac {dy} {dx} = \frac {1} {2}x^\frac {-1} {2}$$

But there's one step that I don't understand and would greatly appreciate if someone could explain it for me. Here it is:

$$y + dy = \sqrt {x} (1 + \frac {dx} {x})^\frac {1} {2}$$

To...

$$= \sqrt {x} + \frac {1} {2} \frac {dx} {\sqrt {x}} - \frac {1} {8} \frac {dx^2} {x \sqrt {x}} +...$$"terms with higher powers of dx"

The last part "terms with higher powers of dx" is exactly how this problem is listed in the book. If anyone at all could explain to me what exactly is going on between these steps I would be so happy!
-CP

Last edited: Aug 21, 2006
2. Aug 21, 2006

### d_leet

It looks like the author is using the binomial theorem, or a taylor series expansion to expand the terms in parentheses raised to the 1/2 power.

3. Aug 21, 2006

### Chemical_Penguin

Any possibility of someone expanding on this? I'm not that familiar with either of those and would like to see how the author can go from step 1 to step 2 in the above equation.

4. Aug 21, 2006

### quasar987

The binomial theorem says nothing about fractional exponents. So he must be using a Taylor expansion, but to find the Taylor expansion of

$$(1 + \frac {dx} {x})^\frac {1} {2}$$

one must know what the derivative of x^½ is! So it doesn't make sense to prove it that way!

Last edited: Aug 21, 2006
5. Aug 21, 2006

### d_leet

I think the binomial theorem can be extended to fractional exponents because I remember this being talked about on the last day of my calc 2 class, but I could be wrong.

6. Aug 21, 2006

### qbert

I think the line of reasoning is:

If y(x) = sqrt(x)
then y(x + dx) = sqrt(x + dx) = sqrt(x)*sqrt(1 + dx/x)

Then do a Taylor's series (or binomial theorem, whichever
says): (1 + z)^(1/2) = 1 + 1/2 z + O(z^2)

So that y(x+dx) - y(x) = sqrt(x) + 1/2 dx/sqrt(x) - sqrt(x) + O(dx^2)

Thus y(x+dx) - y(x) = 1/2 dx/sqrt(x) +O(dx^2)

therefore y'(x) = 1/(2 sqrt(x)).

7. Aug 21, 2006

### quasar987

I'm pretty sure the extension your teacher was refering to is actually just the Taylor series of (1+x)^a ($a\in\mathbb{R}$) which LOOKS like the binomial series:

$$(1+x)^a= 1+\sum_{i=1}^{\infty}\frac{a(a-1)...(a-i+1)}{i!}x^i$$

Notice that the coefficient is the "same"* as the coeffient n!/i!(n-i)! in the binomial series, but with n-->a. But the convergence is for |x|<1 only. However, the point is that we need not know anything about calculus to prove the binomial thm, while we need the knowledge of the derivative of x^a to prove the above formula.

*I mean it follows the natural extension of the notion of factorial from natural to real numbers.

Last edited: Aug 21, 2006
8. Aug 21, 2006

### d_leet

Yea I guess you're right, but we had already finished with taylor series so he introduced this after we talked about the binomial theorem so I guess it just seemed like we were extending the binomial theorem to non integer exponents.

9. Aug 22, 2006

### harish_victory

Hi Friend, I would like to tell u a few words according to my knowledge is concerned. Just have a look at this and if U have any queries don't be late to send me a mail.

We use Taylor's expansion to proove Binomial theorem in which we must know the successive derivatives of "x^a" .
So it doesn't make sense to use again binomial theorem for prooving the derivatives for
x^a where a=1/2etc.
I think - here, to proove these derivatives we can use the concept of

x^n - a^n
Lt ___________ = n*a^(n-1)
x->a x - a
to which I had the proof for all 'n' (fractional & natural).

Use this for prooving the derivatives of x^n where n belongs to real no.s sothat We will be ready with all the derivatives of x^n and hence we can use taylor's theorem for prooving the Binomial theorem.

On the whole I am saying just one thing i.e. it is not right to use binomial or taylor's expansion for prooving the derivatives of x^n.

Last edited: Aug 22, 2006
10. Aug 22, 2006

### Chemical_Penguin

Thanks a bunch for all of the replies guys!

Although I'm still not sure why the author used that as an example as it just caused confusion.

I've decided to just move on in the book and try to not worry about that problem.

Again, thanks for all the help!

11. Aug 23, 2006

### StatusX

A less circular proof might go like this:

$$\frac{d}{dx}(\sqrt{x})=\lim_{h \rightarrow 0} \frac{ \sqrt{x+h}-\sqrt{x}}{h}$$

Now multiply the top and bottom by the conjugate of the top:

$$\left(\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \right) \frac{ \sqrt{x+h}-\sqrt{x}}{h}=\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}$$

So we have:

$$\frac{d}{dx}(\sqrt{x})=\lim_{h \rightarrow 0} \frac{h}{h(\sqrt{x+h}+\sqrt{x})}=\frac{1}{2\sqrt{x}}$$

12. Aug 23, 2006

### quasar987

THAT's the sensible way to do it!

13. Nov 5, 2007

### Alkirin

Question.

I just wrote up an equation and subsequent proof by induction for the derivitive of x^1/2 for any (n+1)th derivitive.

That is, you can plug in any number for N (say, 5) and get the Nth derivitive for x^1/2

My question is...has this already been published? If not, I would like to publish it.

14. Nov 11, 2007

### Gib Z

Yes it is a well known fact that the n-th derivative of a function (in fact, a general function, not just the square root of x) can be expressed quite easily. In fact, the same thing has been done for integration as well, and has lead to the development of fractional calculus, where one finds the "pi-th" derivative of a function, as an example.

15. May 4, 2008

### Isak BM

THis does not make sence as it = zero!

When dividing h by [h(x)]: taking the limit as h goes to zero gives 0/0 which is not defined!

It was a nice try! )

16. May 4, 2008

### d_leet

Taking the limit as h goes to zero is not equivalent to evaluating the expression at h=0.

17. May 4, 2008

### Isak BM

My mistake :P

I see now that h and h cancel and you get 1/x ). Very Nice!!!!

18. May 4, 2008

### lurflurf

It is plenty right, provided you did not use that derivative in proving the binomial or taylor expansion. Personally I would define
x^a=exp(a*log(x))
then
(x^a)'=exp'(a*log(x))*a*log'(x)
(x^a)'=exp(a*log(x))*a/x
(x^a)'=a*x^a/x=a*x^(a-1)
x>0 which we can piece together yielding
(x^a)'==a*x^(a-1)
other wise if we knew
(x^a)'=a*x^(a-1)
and wish to show
(x^(1/a))'=x^(1/a-1)/a
use inverse functions
ie
x=f(g(x))
1=x'=f'(g(x))g'(x)
g'(x)=1/f'(g(x))

19. May 5, 2008

### Feldoh

Yeah but the pi-th derivative doesn't really have any meaning does it? It's cool the do half-derivatives and pith-derivatives, etc., but they really don't have a meaning I don't think?

20. May 5, 2008

### HallsofIvy

Staff Emeritus
Actually, fractional derivatives do have a specific meaning. It's just not normally taught in the basic calculus sequence since it basically involves "operator" algebra.