# Homework Help: Derivative laws question numerical analysis

1. Apr 6, 2010

### nhrock3

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. Apr 6, 2010

### Staff: Mentor

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.
$$\prod_{m = 1}^5 m = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 5! = 120$$

3. Apr 6, 2010

### nhrock3

ok but why the derivative change epsilon to pi?

4. Apr 6, 2010

### Staff: Mentor

What are you talking about? There is no epsilon.

5. Apr 6, 2010

### 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.

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

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