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Derivative laws question numerical analysis

Discussion in 'Calculus and Beyond Homework' started by nhrock3, Apr 6, 2010.

  1. Apr 6, 2010 #1

    i know the law (fg)'=f'g+fg'
    but here there is epsilon
    and it turns to someother sign i dont know what it says.

    i cant understand the transition
  2. jcsd
  3. Apr 6, 2010 #2


    Staff: Mentor

    There's no epsilon in what you posted. Are you talking about [itex]\prod[/itex]? Note that this is upper-case pi, which is different from the constant [itex]\pi[/itex], lower-case pi.

    Upper-case pi is similar to upper-case sigma [itex]\sigma[/itex], except that it is used for products rather than sums. here's an example using the product notation.
    [tex]\prod_{m = 1}^5 m = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 5! = 120[/tex]
  4. Apr 6, 2010 #3
    ok but why the derivative change epsilon to pi?
  5. Apr 6, 2010 #4


    Staff: Mentor

    What are you talking about? There is no epsilon.
  6. Apr 6, 2010 #5


    User Avatar
    Homework Helper

    With epsilon do you mean the summation sign? If so that one is called capital sigma.

    Taking the derivative of a function that is the product of two other functions yields two terms as you know, (fg)'=f'g+g'f. In our case we don't have a function consisting of the product of two functions but of a product of m functions.

    \prod_{j=0}^m f_j=f_0f_1f_2 \cdot ... \cdot f_m

    Now taking the derivative of that product yields:

    \frac{d}{dx}\prod_{j=0}^m=\frac{d}{dx}(f_0f_1f_2 \cdot ... \cdot f_m)=f'_0f_1f_2 \cdot ... \cdot f_m+f_0f_1'f_2 \cdot ... \cdot f_m+...+f_0f_1f_2 \cdot ... \cdot f_m'

    You can see that we get m terms that are added together and every term is the product of m terms. We can write this compactly as:

    \sum_{i=0}^m \frac{df_i}{dx}\prod_{\substack{ j=0 \\ j \neq i}}^m f_j
    Last edited: Apr 6, 2010
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