Derivative laws question numerical analysis

[nhrock3]

i know the law (fg)'=f'g+fg'
but here there is epsilon
and it turns to someother sign i don't know what it says.

i can't understand the transition
?

 

[Mark44]

i know the law (fg)'=f'g+fg'
but here there is epsilon
and it turns to someother sign i don't know what it says.

i can't understand the transition
?

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

Upper-case pi is similar to upper-case sigma \sigma, except that it is used for products rather than sums. here's an example using the product notation.
\prod_{m = 1}^5 m = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 5! = 120

 

[nhrock3]

ok but why the derivative change epsilon to pi?

 

[Mark44]

What are you talking about? There is no epsilon.

 

[Cyosis]

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.

<br /> \prod_{j=0}^m f_j=f_0f_1f_2 \cdot ... \cdot f_m<br />

Now taking the derivative of that product yields:

<br /> \frac{d}{dx}\prod_{j=0}^m=\frac{d}{dx}(f_0f_1f_2 \cdot ... \cdot f_m)=f&#039;_0f_1f_2 \cdot ... \cdot f_m+f_0f_1&#039;f_2 \cdot ... \cdot f_m+...+f_0f_1f_2 \cdot ... \cdot f_m&#039;<br />

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:

<br /> \sum_{i=0}^m \frac{df_i}{dx}\prod_{\substack{ j=0 \\ j \neq i}}^m f_j<br />.