The Leibnitz Formula

Hi,
I need help understanding the Leibnitz formula, I tried a couple of things (see photo) but I can't get a quick way to a high derivative with it?
I think specificly I don't get what the k does in the binomial Co efficient and how to use that to get the answer but I do see where it comes from as I can see the pascals triangle pattern.
Any help is appreciated :smile:

Thanks

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Reply 1

Physics4Life

Like for example I tried to find the 9th derivative of (x^3)(e^2x) but couldn't :3

Reply 2

ztibor

Original post by Physics4Life

Like for example I tried to find the 9th derivative of (x^3)(e^2x) but couldn't :3

The formula for (u*v)

\displaystyle \frac{d^n}{dx^n}\left (u\cdot v\right )=\sum_{k=0}^n \binom {n}{k} \frac{d^{n-k}u}{dx^{n-k}} \cdot \frac{d^kv}{dx^k}

note: The 0th derivative of f(x) (

\frac{d^0 f(x)}{dx^0} =f(x)

) is the
function itself.

for x^3*e^(2x)
order.................e^(2x)..................x^3...............binom (n k)
------------------------------------------------------------------------------------------
0 .................... e^2x .................... x^3 ................1
1.................... 2e^(2x)................. 3x^2................n
2.....................2^2 e^(2x).............6x....................n*(n-1)/2
3.....................2^3 e^(2x).............6.....................n*(n-1)*(n-2)/6
4....................2^4 e^2x................0.....................n*(n-1)*(n-2)(n-3)/4!
...................................................................................................
n....................2^n e^(2x).............0....................
-----------------------------------------------------------------------------------------

For n=9

Unparseable latex formula:

\displaystyle \frac{d^9}{dx^9} (e^{2x}\cdot x^3) =e^{2x}\left (8x^3+108x^2+432x+504)

(edited 9 years ago)

Reply 3

DFranklin

I'm not sure if this helps, but let

D_u

mean "differentiate u w.r.t. x" and

D_v

mean "differentiate v w.r.t. x", and then

D_u^k

mean "differentiate u k times w.r.t. x" etc.

Then Leibniz' rule is the same thing as saying

\dfrac{d^n}{dx^n}(uv) = (D_u + D_v)^n (uv)

, where you expand out

(D_u+D_v)^n

using the binomial theorem just as if it was (A+B)^n.

Reply 4

Physics4Life

Thanks guys, that really helped :smile:

Reply 5

newblood

Original post by DFranklin

I'm not sure if this helps, but let

D_u

mean "differentiate u w.r.t. x" and

D_v

mean "differentiate v w.r.t. x", and then

D_u^k

mean "differentiate u k times w.r.t. x" etc.

Then Leibniz' rule is the same thing as saying

\dfrac{d^n}{dx^n}(uv) = (D_u + D_v)^n (uv)

, where you expand out

(D_u+D_v)^n

using the binomial theorem just as if it was (A+B)^n.

How do you conclude this?

\dfrac{d^n}{dx^n}(uv) = (D_u + D_v)^n (uv)

Reply 6

davros

Original post by newblood

How do you conclude this?

\dfrac{d^n}{dx^n}(uv) = (D_u + D_v)^n (uv)

Personally I would just start with the product rule and think about where all the terms are going to come from when you differentiate repeatedly.

If y = uv then
y' = u'v + uv'
y'' = (u'v)' + (uv')' = u''v + u'v' + u'v' + uv'' = u''v + 2u'v' + uv''
etc

Reply 7

DFranklin

Original post by newblood

How do you conclude this?

\dfrac{d^n}{dx^n}(uv) = (D_u + D_v)^n (uv)

What I posted was solely "here's another way of thinking about it", not any kind of proof.

There are lots of ways you can "see that it works", but if you want to formally prove it I think the quickest thing is going to be induction, which to my mind somewhat obscures what's actually going on.