- #1

nhrock3

- 415

- 0

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

?

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- Thread starter nhrock3
- Start date

- #1

nhrock3

- 415

- 0

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

Mark44

Mentor

- 36,308

- 8,280

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.

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

?

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]

- #3

nhrock3

- 415

- 0

ok but why the derivative change epsilon to pi?

- #4

Mark44

Mentor

- 36,308

- 8,280

What are you talking about? There is no epsilon.

- #5

Cyosis

Homework Helper

- 1,495

- 0

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.

[tex]

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

[/tex]

Now taking the derivative of that product yields:

[tex]

\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'

[/tex]

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:

[tex]

\sum_{i=0}^m \frac{df_i}{dx}\prod_{\substack{ j=0 \\ j \neq i}}^m f_j

[/tex].

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.

[tex]

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

[/tex]

Now taking the derivative of that product yields:

[tex]

\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'

[/tex]

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:

[tex]

\sum_{i=0}^m \frac{df_i}{dx}\prod_{\substack{ j=0 \\ j \neq i}}^m f_j

[/tex].

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