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Leibniz Rule (Differentiation)

Can someone point me in the right direction on how to go about this question?

Consider y=ex22y=e^{\frac{x^2}{2}}. Show that dydx=xy\frac{dy}{dx}=xy - I can do that.

By differentiating this equation n times using Leibniz's rule, show that y(n+1)(x)=xy(n)(x)+ny(n1)(x)y^{(n+1)}(x)=xy^{(n)}(x)+ny^{(n-1)}(x).

I literally have no idea where to start =\ .
Original post by ViralRiver
Can someone point me in the right direction on how to go about this question?

Consider y=ex22y=e^{\frac{x^2}{2}}. Show that dydx=xy\frac{dy}{dx}=xy - I can do that.

By differentiating this equation n times using Leibniz's rule, show that y(n+1)(x)=xy(n)(x)+ny(n1)(x)y^{(n+1)}(x)=xy^{(n)}(x)+ny^{(n-1)}(x).

I literally have no idea where to start =\ .

Without sounding harsh, have you actually tried to plug it into the General Leibniz rule?
(namely dndxn(f(x)g(x))=k=0n(nk)(dkdxk[f(x)])(dnkdxnk[g(x)])\dfrac{d^n}{dx^n}\left(f(x)\cdot g(x) \right) = \displaystyle\sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \left(\dfrac{d^k}{dx^k}[f(x)]\right) \left(\dfrac{d^{n-k}}{dx^{n-k}}[g (x)]\right) with f(x)=xf(x)=x and g(x)=y?)g(x)=y?)
(edited 12 years ago)
Reply 2
Original post by Farhan.Hanif93
Without sounding harsh, have you actually tried to plug it into the General Leibniz rule?
(namely dndxn(f(x)g(x))=k=0n(dkdxk[f(x)])(dnkdxnk[g(x)])\dfrac{d^n}{dx^n}\left(f(x)\cdot g(x) \right) = \displaystyle\sum_{k=0}^n \left(\dfrac{d^k}{dx^k}[f(x)]\right) \left(\dfrac{d^{n-k}}{dx^{n-k}}[g (x)]\right) with f(x)=xf(x)=x and g(x)=y?)g(x)=y?)


That's not how I learned it. The version I have learned is:

Dn(uv)=(Dnu)v+(n1)(Dn1u)(Dv)++(nr)(Dnru)(Drv)++u(Drv)D^n(uv)=(D^nu)v+ \left( {\begin{array}{cc} n \\ 1 \\ \end{array} } \right) (D^{n-1}u)(Dv)+ \cdot \cdot \cdot+\left( {\begin{array}{cc} n \\ r \\ \end{array} } \right) (D^{n-r}u)(D^rv) + \cdot \cdot \cdot + u(D^rv).

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

Dn(uv)=(Dnu)v+(n1)(Dn1u)(Dv)++(nr)(Dnru)(Drv)++u(Drv)D^n(uv)=(D^nu)v+ \left( {\begin{array}{cc} n \\ 1 \\ \end{array} } \right) (D^{n-1}u)(Dv)+ \cdot \cdot \cdot+\left( {\begin{array}{cc} n \\ r \\ \end{array} } \right) (D^{n-r}u)(D^rv) + \cdot \cdot \cdot + u(D^rv).

I've tried substituting into that, but it's a tad confusing.

I've never come across that notation before but I assume that Dn(y)=dnydxnD^n(y) = \dfrac{d^ny}{dx^n}? In which case, you should be able to see that both versions are equivalent. In your notation, take v=xv=x and u=yu=y. Notice that all terms have a factor of Dr(x)D^{r}(x). What happens to that quantity when r2?r\geq 2? (if it's not immediate, try a few values.)
Reply 4
Original post by Farhan.Hanif93
I've never come across that notation before but I assume that Dn(y)=dnydxnD^n(y) = \dfrac{d^ny}{dx^n}? In which case, you should be able to see that both versions are equivalent. In your notation, take v=xv=x and u=yu=y. Notice that all terms have a factor of Dr(x)D^{r}(x). What happens to that quantity when r2?r\geq 2? (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?
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 (nr)\begin{pmatrix} n \\ r \end{pmatrix} coefficients... :sigh: Sorry for any confusion on that front. :o:
Reply 6
dndxny=dndxn(xy)=k=0n(nCk)(dkdxk[x])(dnkdxnk[y])\dfrac{d^n}{dx^n}y' = \dfrac{d^n}{dx^n}\left( x\cdot y \right) = \displaystyle\sum_{k=0}^n \left( {nCk}\right) \left(\dfrac{d^k}{dx^k}[x]\right) \left(\dfrac{d^{n-k}}{dx^{n-k}}[y]\right)

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


(nC0)x<0>y<n>+(nC1)x<1>y<n1>+ k=2n(nCk)(dkdxk[x])(dnkdxnk[y])\left( {nC0}\right)x^{<0>}y^{<n>}+ \left( {nC1}\right) x^{<1>}y^{<n-1>} + \ \displaystyle\sum_{k=2}^{n} \left( {nCk}\right) \left(\dfrac{d^k}{dx^k}[x]\right) \left(\dfrac{d^{n-k}}{dx^{n-k}}[y]\right)

get the binomial coefficients and differentiate the second x term once

(1)xy<n>+(n)1y<n1>+0\left( {1}\right)xy^{<n>}+ \left( {n}\right) 1y^{<n-1>} + 0

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 y<n+1>=xy<n>+ny<n1> y^{<n+1>} = xy^{<n>} + ny^{<n-1>}
(edited 12 years ago)
Reply 7
Thanks a lot for your help guys :smile: - makes a lot more sense now.
Reply 8
Original post by swelshie
dndxny=dndxn(xy)=k=0n(nCk)(dkdxk[x])(dnkdxnk[y])\dfrac{d^n}{dx^n}y' = \dfrac{d^n}{dx^n}\left( x\cdot y \right) = \displaystyle\sum_{k=0}^n \left( {nCk}\right) \left(\dfrac{d^k}{dx^k}[x]\right) \left(\dfrac{d^{n-k}}{dx^{n-k}}[y]\right)

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


Unparseable latex formula:

\left( {nC0}\right)x^{&lt;0&gt;}y^{&lt;n&gt;}+ \left( {nC1}\right) x^{&lt;1&gt;}y^{&lt;n-1&gt;} + \ \displaystyle\sum_{k=2}^{n} \left( {nCk}\right) \left(\dfrac{d^k}{dx^k}[x]\right) \left(\dfrac{d^{n-k}}{dx^{n-k}}[y]\right)



get the binomial coefficients and differentiate the second x term once

Unparseable latex formula:

\left( {1}\right)xy^{&lt;n&gt;}+ \left( {n}\right) 1y^{&lt;n-1&gt;} + 0



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
Unparseable latex formula:

y^{&lt;n+1&gt;} = xy^{&lt;n&gt;} + ny^{&lt;n-1&gt;}




Original post by Farhan.Hanif93
You're right, it is late! I even forgot to put in the (nr)\begin{pmatrix} n \\ r \end{pmatrix} coefficients... :sigh: Sorry for any confusion on that front. :o:


Ok it now asks me to evaluate y(5)(0)y^{(5)}(0). My guess: y(5)(0)=xy(4)(0)+4y(3)(0)=4y(3)(0)y^{(5)}(0)=xy^{(4)}(0)+4y^{(3)}(0)=4y^{(3)}(0) but not really sure what to do from there (the answer should be 0).

EDIT: I've put y(3)(0)y^{(3)}(0) into the Leibniz thing and got it equal to 2y' which is of course 0, hence y(5)(0)=0y^{(5)}(0)=0 - is that the right method?
(edited 12 years ago)
Original post by ViralRiver
Ok it now asks me to evaluate y(5)(0)y^{(5)}(0). My guess: y(5)(0)=xy(4)(0)+4y(3)(0)=4y(3)(0)y^{(5)}(0)=xy^{(4)}(0)+4y^{(3)}(0)=4y^{(3)}(0) but not really sure what to do from there (the answer should be 0).

EDIT: I've put y(3)(0)y^{(3)}(0) into the Leibniz thing and got it equal to 2y' which is of course 0, hence y(5)(0)=0y^{(5)}(0)=0 - is that the right method?

Yep, that's fine.

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