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Math Help - Help needed Thank You in Advance

  1. #1
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    Talking Help needed Thank You in Advance

    Question: Using binomial coefficients, derive, a formula for the Nth derivative of the product of two functions.

    Thank You

    Dave
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  2. #2
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    y(x)=f(x)g(x)
    y'(x)=f'(x)g(x)+f(x)g'(x)
    y''(x)=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)= \left(\begin{array}{c}2\\0\end{array}\right)f''(x)  g(x)+\left(\begin{array}{c}2\\1\end{array}\right)f  '(x)g'(x)+\left(\begin{array}{c}2\\2\end{array}\ri  ght)f(x)g''(x)
    y^{(3)}(x)=f^{(3)}(x)g(x)+3f''(x)g'(x)+3f'(x)g''(x  )+f(x)g^{(3)}(x)= \left(\begin{array}{c}3\\0\end{array}\right)f^{(3)  }(x)g(x)+\left(\begin{array}{c}3\\1\end{array}\rig  ht)f''(x)g'(x)+\left(\begin{array}{c}3\\3\end{arra  y}\right)f'(x)g''(x)+\left(\begin{array}{c}2\\0\en  d{array}\right)f(x)g^{(3)}(x)

    Can you continue?

    Leibniz rule (generalized product rule) - Wikipedia, the free encyclopedia
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  3. #3
    MHF Contributor

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    You can prove the general formula by induction:

    (f^{n+1}) ' = (f^{n} f) ' = (f^n) ' f + f^n f '.
    Last edited by mr fantastic; May 15th 2009 at 04:10 AM. Reason: Added [/math]
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  4. #4
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    sorry i dont understand how to get the formula after that, i understand the use of the product rule and the binomial thoerem, but i dont understand wer to go after that
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