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Leibniz Rule (Differentiation)

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

    Consider y=e^{\frac{x^2}{2}}. Show that \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^{(n-1)}(x).

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

    Consider y=e^{\frac{x^2}{2}}. Show that \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^{(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 \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)=x and g(x)=y?)
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    (Original post by Farhan.Hanif93)
    Without sounding harsh, have you actually tried to plug it into the General Leibniz rule?
    (namely \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)=x and g(x)=y?)
    That's not how I learned it. The version I have learned is:

    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.
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    (Original post by ViralRiver)
    That's not how I learned it. The version I have learned is:

    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 D^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=x and u=y. Notice that all terms have a factor of D^{r}(x). What happens to that quantity when r\geq 2? (if it's not immediate, try a few values.)
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    (Original post by Farhan.Hanif93)
    I've never come across that notation before but I assume that D^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=x and u=y. Notice that all terms have a factor of D^{r}(x). What happens to that quantity when 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?
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    (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 \begin{pmatrix} n \\ r \end{pmatrix} coefficients... :sigh: Sorry for any confusion on that front. :o:
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    \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.


    \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

    \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^{<n-1>}
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    Thanks a lot for your help guys - makes a lot more sense now.
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    (Original post by swelshie)
    \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.


    \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

    \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^{<n-1>}

    (Original post by Farhan.Hanif93)
    You're right, it is late! I even forgot to put in the \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). My guess: 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) into the Leibniz thing and got it equal to 2y' which is of course 0, hence y^{(5)}(0)=0 - is that the right method?
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    (Original post by ViralRiver)
    Ok it now asks me to evaluate y^{(5)}(0). My guess: 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) into the Leibniz thing and got it equal to 2y' which is of course 0, hence y^{(5)}(0)=0 - is that the right method?
    Yep, that's fine.

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Updated: November 19, 2011
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