# Homework Help: Leibintz's theorem

1. Jan 17, 2007

### gohar

hi all im learner...i want to know leibintz's theorem
need help from experts.......

2. Jan 17, 2007

### cristo

Staff Emeritus
What specifically are you having trouble with? What do you know? You'll have to be more specific in order to get help!

3. Jan 17, 2007

### HallsofIvy

Do you mean this one:
$$\frac{d}{dx}\int_{\alpha (x)}^{\beta (x)}\Phi(x,t)dt= \int_{\alpha(x)}^{\beta(x)}\frac{\partial \Phi(x,t)}{\partial x}dt+\frac{d\beta (x)}{dx}\Phi(x,\beta (x)) - \frac{d\alpha (x)}{dx}\Phi(x,\alpha (x))$$

Do you want just the statement or do you have a question about it?

4. Jan 19, 2007

### gohar

no i want someone to explane leibintz's theorem for finding nth derivative i could'nt understand in class.. {UV} some thing..

5. Jan 19, 2007

### HallsofIvy

Perhaps you mean this:

$$\frac{d^n (f(x)g(x))}{dx^n}= \Sigma_{i=0}^n \left(\begin{array}{c}n \\ i\end{array}\right)f^{(i)}g^{(n-i)}$$
Where $$\left(\begin{array}{c}n \\ i\end{array}\right)$$ is the "binomial coefficient".

Do you understand why the "binomial theorem" $$(x+ y)^n= \Sigma_{i=0}^n \left(\begin{array}{c}n \\ i\end{array}\right)x^iy^{n-i}$$ works? It is much the same thing.

(I googled on "Leibniz's theorem" and got about a dozen different formulas!)

6. Jan 19, 2007

### Gib Z

Leibniz was a smart guy :)

I prefer using the nCr notation, for some reasons it makes it more obvious to me very his theorem came from. The 'binomial coefficent' make me think he did some fancy stuff with Binomial theorem, when really its an easy relationship from the definition of nCr.

Comes from the fact then when we derive x^n repeatedly, the coefficents form into n(n-1)(n-2)(n-3)...look familiar :) I didnt help much lol

O wait I didnt answer anything the OP wanted.

Ok, lets try a simple function, like the one I gave above. $$f(x)=x^n$$ Then $$f'(x)=nx^{n-1}$$. Again,
$$f''(x)=n(n-1)x^{n-2}$$. More: $$f^3(x)=n(n-1)(n-2)x^{n-3}$$. Can you start to see the pattern. With the i-th derivative, the power is n-i, and the co efficent is (n!)/(n-i+1)!.

The theorem generalises this into any f(x).

Holy wack, You know what I just realised.

Correct me if im wrong, because I probably am, But...Using the Gamma Function, could we find non-integer derivatives? I can't imagine those in my head...please tell me that I could be onto something and no bodys done it before, maybe I could do a thesis on this :)

Last edited: Jan 19, 2007
7. Jan 20, 2007

### HallsofIvy

Here is one way to think about the "binomial coefficient" and why it also pops up in the extended product rule for differentiation.

Suppose we were to multiply (x+ y)3 without combining terms. Also I am going to write xxx instead of x3 which, I think, will make my point clearer.

(x+y)2= xx+ xy+ yx+ yy so
(x+y)3= (x+y)(xx+ xy+ yx+ yy)= xxx+xxy+xyx+xyy+ yxx+yxy+yyx+yyy.

Now, because there are 3C1= 3 ways of writing "two x's and one y" (xxy, xyx, yxx) as well as 3C2= 3 ways of writing "one x and two y's" (xyy, yxy, yyx), that is x3+ 3x2y+ 3xy2+ x3.

Now do (fg)''' the same way: by the product rule (fg)'= f'g+ fg'. (fg)''= (f'g+ fg')'= f''g+ f'g'+ f'g'+ fg''. (fg)'''= (f''g+ f'g'+ f'g'+ fg'')= f'''g+ f''g'+ f''g'+ f'g''+ f'g''+ fg'''. Again, it is because the binomial coefficient answers the question "How many ways can we write n letters if i of the are the same (x or f) and the other n-i are the same (y or g) that the binomial coefficient occurs in both (x+ y)n and (fg)(n).

8. Jan 20, 2007

### slider142

Good insight! This is the start of fractional calculus. See http://en.wikipedia.org/wiki/Fractional_derivative .

9. Jan 20, 2007

### gohar

Thanks all friends........

Thanks all friends........
The nth derivative of 1/1+x is (-1)^n *n! /(1+x)^n+1
& i think the nth derivative of cosx = cos (x + (n*pi)/2)
how to find the nth derivative ?
one is the leibintz's theorem...

10. Jan 20, 2007

### Gib Z

Your joking, I was right? Aww but looking at that link it looks like im too late lol.

gohar...sometimes its easier just to search for a pattern. Its simple enough to see $$f(x)=kx^n$$ then $$f^z(x)=nP_z \cdot kx^{n-z}$$. Or, even easier, $$f(x)=k^x$$ then $$f^n(x)=(log_e k)^n \cdot k^x$$. If your looking for the most general case, Then Leibniz is your friend.

11. Jan 21, 2007

### gohar

Gib Z u r right but i want to know how to differentiate a function by leibniz therom.....?

12. Jan 21, 2007

### cristo

Staff Emeritus
The expression given to you in post #5 by HallsofIvy tells you how to find the nth derivative of a function.