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Chain rule with second derivative

  1. Aug 20, 2010 #1
    1. The problem statement, all variables and given/known data

    I trying to find the second derivative of [itex]xe^x[/itex]

    2. Relevant equations

    chain rule

    3. The attempt at a solution

    Two find the first derivative I use the chain rule.

    [itex]f'(y)g(y)+f(y)g'(y)[/itex]

    so I get

    [itex]e^x+xe^x[/itex]

    is the second derivative

    [itex]e^x+f'(y)g(y)+f(y)g'(y)[/itex]

    = [itex]e^x+e^x+xe^x[/itex]
    = [itex]2e^x+xe^x[/itex]

    regards
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 20, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    That should be correct, also, you used the product rule, not the chain rule.
     
  4. Aug 21, 2010 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It can be shown that, for any positive integer n, with [itex]f^{(n)}(x)[/itex] indicating the nth derivative, that
    [tex](fg)^{(n)}(x)= \sum_{i=0}^n \begin{pmatrix}n \\ i\end{pmatrix}f^{(i)}(x)g^{(n- i)}(x)[/tex]

    And, as rock.freak667 said, that is the product rule.
     
  5. Aug 22, 2010 #4
    thanks guys
     
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