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Math Help - find the n'th differentiation of these..

  1. #1
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    find the n'th differentiation of these..

    1=<k=<n
    find  f^{(n)} (x) of:

    the first :
    <br />
   f(x) = |x|^{n + 1}<br />
    the second is:
        <br />
  f(x) = |\sin x|^{n + 1}   <br />
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  2. #2
    Member Abu-Khalil's Avatar
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    The first function k-derivate is f^{k}(x)=\begin{cases}\displaystyle\frac{(n+1)!}{(  n+1-k)!}x^{n+1-k},x>0\\ \\  (-1)^{n+1}\displaystyle\frac{(n+1)!}{(n+1-k)!}x^{n+1-k},x<0\\\end{cases}.

    The second involves a lots of fun because you'll have to do the same with a product !
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  3. #3
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    what is the general way you found this result??

    what is the derivative of absolute value function??
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  4. #4
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    how you find the n'th derivative
    did you use macloren series formula
    ??
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  5. #5
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    Quote Originally Posted by Abu-Khalil View Post
    The first function k-derivate is f^{k}(x)=\begin{cases}\displaystyle\frac{(n+1)!}{(  n+1-k)!}x^{n+1-k},x>0\\ \\ (-1)^{n+1}\displaystyle\frac{(n+1)!}{(n+1-k)!}x^{n+1-k},x<0\\\end{cases}.

    The second involves a lots of fun because you'll have to do the same with a product !
    how you got this solution??
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  6. #6
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    when i use leibnitz formula

    for the second question:
    <br />
(|\sin x|^{n}||\sin x|)^{(n)}=\sum_{k=0}^{n}C^k_n (|\sin x|^{n})^{(n-k)}|sinx|^{(k)}\\<br />

    what to do next??
    Last edited by transgalactic; February 9th 2009 at 07:49 AM.
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  7. #7
    Member Abu-Khalil's Avatar
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    Quote Originally Posted by transgalactic View Post
    how you got this solution??
    Just observing what happened every time i derivated the function and tried to make a rule.
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  8. #8
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    what about the solution i proposed??
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  9. #9
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    Quote Originally Posted by transgalactic View Post
    its similar to
    when i use leibnitz formula
    in this case g(x)=1
    <br />
(fg)^{(n)}=\sum_{k=0}^{n}C^k_n |x|^{{n + 1}^{(n-k)}g^{(k)}}<br />
    so by this solution
    i just need to input f(x) there

    but it differs your solution??
    is this a correct way??
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  10. #10
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    Quote Originally Posted by Abu-Khalil View Post
    The first function k-derivate is f^{k}(x)=\begin{cases}\displaystyle\frac{(n+1)!}{(  n+1-k)!}x^{n+1-k},x>0\\ \\  (-1)^{n+1}\displaystyle\frac{(n+1)!}{(n+1-k)!}x^{n+1-k},x<0\\\end{cases}.

    The second involves a lots of fun because you'll have to do the same with a product !
    i need the n'th derivative

    not the k'th in terms of n

    how to change that??
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