Derivative proof with a fractional exponent

[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!
-Thanks in advance.
-CP

[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.

[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.

[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!

[d_leet]

The binomial theorem says nothing about fractional exponents.


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.

[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)).

[quasar987]

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.

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.

[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.

[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.

[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.:biggrin:

Again, thanks for all the help!

[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 }}

[quasar987]

THAT's the sensible way to do it! :biggrin:

[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.

[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.

[Isak BM]

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 }}

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! :o)

[d_leet]

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! :o)

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

[Isak BM]

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

[lurflurf]

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.
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))

[Feldoh]

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.

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?

[HallsofIvy]

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.

[Big-T]

Would you care to elaborate on that, HallsofIvy?

[Redbelly98]

I have no problem with the concept of fractional derivatives (they can be defined using Fourier transforms), but do they have an application somewhere?

[Feldoh]

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.

Yeah I'm kind of curious as to their meaning... All you really seem to need to know in terms of algebra is the gamma function isn't it?

[PowerIso]

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

By the way, he is using the definition of a derivative, so it better work out!

[HallsofIvy]

I'm no expert on the subject of fractional derivatives so I googled it:
http://en.wikipedia.org/wiki/Fractional_calculus
http://mathworld.wolfram.com/FractionalDerivative.html
http://mathworld.wolfram.com/FractionalDerivative.html

[Feldoh]

Actually I though about the whole fraction derivative and I've hit a bump.

The first derivative is defined as

f'(x)= \lim_{\substack{h\rightarrow 0}}\frac{f(x+h)- f(x)}{h}

So how would one define, say, the 1/2 derivative?

[Redbelly98]

... how would one define, say, the 1/2 derivative?

It can be done using Fourier transforms, if you're familiar with them.

Consider this function:

f(x) = Ae^{i k x}

where A is a constant.

To take the nth derivative of this function, just multiply it by a factor (ik)^n so that


\frac{d^n}{dx^n} f(x) = (i k)^n f(x)


So for this particular function, it is easy to define a non-integer derivative. Just let n = 1/2, for example.

If you're not familar with Fourier transforms, just be aware that any, arbitrary function f(x) can be expressed in the form:


f(x) = \int F(k) e^{i k x} dk


where the integration limits are -\infty to +\infty, and the function F(k) uniquely determines f(x).

To take the n-th derivative, multiply by (ik)^n within the integrand (similar to what we did above):


\frac{d^n}{dx^n} f(x) = \int F(k) (i k)^n e^{i k x} dk


Again, n can be 1/2 or anything you want.

[Pere Callahan]

Have a look at the links provided by HallsofIvy.
Fractional derivatives are not defined as the limit of some difference quotient.

[Feldoh]

Have a look at the links provided by HallsofIvy.
Fractional derivatives are not defined as the limit of some difference quotient.

Yeah, unfortunately the only one I can actually follow is http://en.wikipedia.org/wiki/Fractional_calculus#Half_derivative_of_a_simple_fu nction

Which bases that concept of a fractional derivative based off of the the definition of derivative.

[HallsofIvy]

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! :o)

I have absolutely no idea what you mean by "dividing h by [h(x)]" since there was no "h(x)" mentioned.

But have you noticed that in the general formula for the derivative,
\lim_{\stack{h\rightarrow 0}} \frac{f(x+h)- f(x)}{h}
it is always of the form "0/0"? Are you saying that the whole concept of "derivative" is a "nice try"?

The fact that substituting h= 0 in the formula gives 0/0 does NOT mean the limit itself does not exist. That should have been one of the first things you learned in Calculus.

[mathwonk]

1) the binomial theorem for fractional exponents is due to newton.

2) if you want very rigorous proofs of every detail in calculus, you should not be reading calculus made easy, which is an intuitive presentation.

3) fractional derivatives are treated by riemann, in the complex case, by the cauchy integral formula, which changes the order of the derivative, into the order of an exponent under the integral sign.

i.e. since one can integrate fractional exponents, one can take fractional derivatives.

[DavidWhitbeck]

The binomial theorem says nothing about fractional exponents.

That is simply not true. Just use the gamma function in place of factorials, and the rest is the same.

[mathwonk]

The difference is semantic, since people who know only a little math, may think the phrase "binomial theorem" refers only to the one they learnt for integer powers.

[Vid]

http://www.amazon.com/Applications-Fractional-Calculus-Physics-Rudolf/dp/9810234570

For anyone interested especially with amazon's nifty search inside feature.

[Vid]

There's also a $10 dover book for those really interested.

http://www.amazon.com/Fractional-Calculus-Applications-Differentiation-Integration/dp/0486450015/ref=pd_sim_b_title_1

[robert Ihnot]

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

There is some confusion with the above as written since higher powers of dx refer, for example, to (dx)^n, which is to say we are only taking exponents, not derivatives.

What is involved is sometimes (Wikipedia) referred to as Newton's Generalized Binominal Theorem.

[robert Ihnot]

Vid: There's also a $10 dover book for those really interested.

Thanks! That is certainly interesting. I can remember thinking about those things, but I did not recognize there was so much written on it.

[Kurret]

My proof would look like this, which i think is easy to understand:
first the proof for y=x^2
f(x)=x^2
f'(x)= \lim_{\substack{h\rightarrow 0}}\frac{(x+h)^2- x^2}{h}=\lim_{\substack{h\rightarrow 0}}\frac{2xh+h^2}{h}=\lim_{\substack{h\rightarrow 0}}2x+h=2x
now let f(x)=x^{\frac{1}{2}}
equivalent to:
f(x)^2=x
differentiate both side with respect to x, and using chainrule and the statement for x^2 we proved earlier:
2f(x)\cdot f'(x)=1
f'(x)=\frac{1}{2f(x)}=\frac{1}{2x^{\frac{1}{2}}}
this proof can easily be generalized to all rational numbers

[Big-T]

Kurret: This is not the concept discussed, you have taken the first derivative of some function. The topic is fractional derivatives (for example: what does it mean to take the pi'th derivative of a function?) not the derivatives of functions of fractional powers.,

[Kurret]

Kurret: This is not the concept discussed, you have taken the first derivative of some function. The topic is fractional derivatives (for example: what does it mean to take the pi'th derivative of a function?) not the derivatives of functions of fractional powers.,
Yes the discussion may have changed to that, but the thread starters question was about the first derivative of x^(1/2).

[Big-T]

I agree, sorry.

[gamesguru]

This is easy, and the way I do it can be applied to any rational exponent.
y=\sqrt{x}
y^2=x
2y \frac{dy}{dx}=1
\frac{dy}{dx}=\frac{1}{2y}=\frac{1}{2\sqrt{x}}.