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# Leibniz Rule (Differentiation)

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1. Leibniz Rule (Differentiation)

Consider . Show that - I can do that.

By differentiating this equation n times using Leibniz's rule, show that .

I literally have no idea where to start =\ .
2. Re: Leibniz Rule (Differentiation)
(Original post by ViralRiver)

Consider . Show that - I can do that.

By differentiating this equation n times using Leibniz's rule, show that .

I literally have no idea where to start =\ .
Without sounding harsh, have you actually tried to plug it into the General Leibniz rule?
(namely with and
Last edited by Farhan.Hanif93; 19-11-2011 at 04:50. Reason: It's late...
3. Re: Leibniz Rule (Differentiation)
(Original post by Farhan.Hanif93)
Without sounding harsh, have you actually tried to plug it into the General Leibniz rule?
(namely with and
That's not how I learned it. The version I have learned is:

.

I've tried substituting into that, but it's a tad confusing.
4. Re: Leibniz Rule (Differentiation)
(Original post by ViralRiver)
That's not how I learned it. The version I have learned is:

.

I've tried substituting into that, but it's a tad confusing.
I've never come across that notation before but I assume that ? In which case, you should be able to see that both versions are equivalent. In your notation, take and . Notice that all terms have a factor of . What happens to that quantity when (if it's not immediate, try a few values.)
5. Re: Leibniz Rule (Differentiation)
(Original post by Farhan.Hanif93)
I've never come across that notation before but I assume that ? In which case, you should be able to see that both versions are equivalent. In your notation, take and . Notice that all terms have a factor of . What happens to that quantity when (if it's not immediate, try a few values.)
I'll take another look at it tomorrow - don't think I can work this late :P . However, I don't see any form of binomials in your expression?
6. Re: Leibniz Rule (Differentiation)
(Original post by ViralRiver)
I'll take another look at it tomorrow - don't think I can work this late :P . However, I don't see any form of binomials in your expression?
You're right, it is late! I even forgot to put in the coefficients... Sorry for any confusion on that front.
7. Re: Leibniz Rule (Differentiation)

evaluate first two terms in the sum ie k = 0 and 1.

get the binomial coefficients and differentiate the second x term once

the rest of the sum is zero as all remaining derivatives of x order 2 or higher are zero (differentiation with respect to x)
giving
Last edited by swelshie; 19-11-2011 at 08:15.
8. Re: Leibniz Rule (Differentiation)
Thanks a lot for your help guys - makes a lot more sense now.
9. Re: Leibniz Rule (Differentiation)
(Original post by swelshie)

evaluate first two terms in the sum ie k = 0 and 1.

get the binomial coefficients and differentiate the second x term once

the rest of the sum is zero as all remaining derivatives of x order 2 or higher are zero (differentiation with respect to x)
giving

(Original post by Farhan.Hanif93)
You're right, it is late! I even forgot to put in the coefficients... Sorry for any confusion on that front.
Ok it now asks me to evaluate . My guess: but not really sure what to do from there (the answer should be 0).

EDIT: I've put into the Leibniz thing and got it equal to 2y' which is of course 0, hence - is that the right method?
Last edited by ViralRiver; 19-11-2011 at 22:59.
10. Re: Leibniz Rule (Differentiation)
(Original post by ViralRiver)
Ok it now asks me to evaluate . My guess: but not really sure what to do from there (the answer should be 0).

EDIT: I've put into the Leibniz thing and got it equal to 2y' which is of course 0, hence - is that the right method?
Yep, that's fine.

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Last updated: November 19, 2011
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