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Math Help - n-th derivative of u'exp(u)

  1. #1
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    n-th derivative of u'exp(u)

    Hello...

    BIG question (big for me, at least): Is there a general formula for the n-th derivative of

    <br />
u'e^u\\<br />

    Basically I have a function exactly like that and I need to extract the n-th derivative.

    THANKS in advance
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  2. #2
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    Quote Originally Posted by paolopiace View Post
    Hello...

    BIG question (big for me, at least): Is there a general formula for the n-th derivative of

    <br />
u'e^u\\<br />

    Basically I have a function exactly like that and I need to extract the n-th derivative.

    THANKS in advance
    I noticed a pattern at the 2nd derivative...

    f(u)=u'e^u \Rightarrow u^{(1)}e^u

    f^{(1)}(u)=u^{(2)}e^u+(u^{(1)})^2e^u

    f^{(2)}(u)=u^{(3)}e^u+u^{(2)}u^{(1)}e^u + 2u^{(2)}u^{(1)}e^u+\left(u^{(3)}\right)^2e^u

    f^{(2)}(u)=u^{(3)}e^u+3u^{(2)}u^{(1)}e^u+\left(u^{  (3)}\right)^2e^u

    I may be wrong, but I will assume this:

    f^{(n)}(u)=u^{(n-1)}e^u+(n+1)u^{(n)}u^{(n-1)}\cdots u^{(1)}e^u+\left(u^{(n+1)}\right)^{n-1}e^u
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  3. #3
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    Quote Originally Posted by colby2152 View Post
    I noticed a pattern at the 2nd derivative...

    f(u)=u'e^u \Rightarrow u^{(1)}e^u

    f^{(1)}(u)=u^{(2)}e^u+(u^{(1)})^2e^u

    f^{(2)}(u)=u^{(3)}e^u+u^{(2)}u^{(1)}e^u + 2u^{(2)}u^{(1)}e^u+\left(u^{(3)}\right)^2e^u

    f^{(2)}(u)=u^{(3)}e^u+3u^{(2)}u^{(1)}e^u+\left(u^{  (3)}\right)^2e^u

    I may be wrong, but I will assume this:

    f^{(n)}(u)=u^{(n-1)}e^u+(n+1)u^{(n)}u^{(n-1)}\cdots u^{(1)}e^u+\left(u^{(n+1)}\right)^{n-1}e^u
    There's a formula for taking the nth derivative of any function of the form y = f(x) g(x) called Leibniz's Rule.
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    Thanks. I know the Leibniz rule. I just wanted...

    ...to see if there is a more simple and straightforward rule for u'exp(u).
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    Quote Originally Posted by paolopiace View Post
    Hello...

    BIG question (big for me, at least): Is there a general formula for the n-th derivative of

    <br />
u'e^u\\<br />

    Basically I have a function exactly like that and I need to extract the n-th derivative.

    THANKS in advance
    Since u'e^u is the derivative of e^u, you only need to consider the (n-1)-th derivative of e^u.
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    Quote Originally Posted by mr fantastic View Post
    There's a formula for taking the nth derivative of any function of the form y = f(x) g(x) called Leibniz's Rule.
    Yeah, I never really used that rule. It is handy though - it is basically a binomial expansion, but for derivatives. Did I get the derivative right, or are there more terms in the middle?
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  7. #7
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    Quote Originally Posted by paolopiace View Post
    Is there a general formula for the n-th derivative of

    <br />
u'e^u\\<br />

    Basically I have a function exactly like that and I need to extract the n-th derivative.
    If y = u'e^u then y^{(n)} = D_n(u)e^u, where D_n(u) is some polynomial in u and its derivatives. If you differentiate one more time then you get y^{(n+1)} = (D_n'(u) + u'D_n(u))e^u. So D_{n+1} = D_n'(u) + u'D_n(u). If you apply this recursive formula to find the first few derivatives then you get

    y^{(1)} = y' = \bigl((u^{(1)})^2 + u^{(2)}\bigr)e^u,
    y^{(2)} = \bigl((u^{(1)})^3 + 3u^{(1)}u^{(2)}+ u^{(3)}\bigr)e^u,
    y^{(3)} = \bigl((u^{(1)})^4 + 4u^{(1)}u^{(3)}+ 6(u^{(1)})^2u^{(2)} + 3(u^{(2)})^2 + u^{(4)}\bigr)e^u,
    y^{(4)} = \bigl((u^{(1)})^5 + 10(u^{(1)})^3u^{(2)} + 14(u^{(1)})^2u^{(3)} + 15u^{(1)}(u^{(2)})^2 + 5u^{(1)}u^{(4)} + 6u^{(2)}u^{(3)} + u^{(5)}\bigr)e^u.

    That should be enough to convince you that there is not going to be any simple formula for the n-th derivative.
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