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How do you differentiate one? Is it possible? Any formulas?

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- Thread starter the dude man
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- #1

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How do you differentiate one? Is it possible? Any formulas?

- #2

AKG

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Apply d/dx to the expression.

- #3

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- #4

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y'(x) = y(x)*M

M=

----

\ f' / f

/

----

Does that work?

M=

----

\ f' / f

/

----

Does that work?

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- #5

AKG

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No, that doesn't work. Change one of the f's to a g, then it should work.

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I cant figure this out can someone refer me to proof cause i cant find one.

- #7

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The derivative of a function in the form of an infinite product.

Anything would be nice.

Anything would be nice.

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Bump. .

- #9

matt grime

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- #10

3.14159

A similar question came to me while finding the derivative of a function equal to the product of three differentiable functions here is my generalization to a function equal to an infinite product:

Let [tex] g_{n} = \prod^n_{k=1}f_{k}[/tex] and assume that [tex]f'_n[/tex] exists. Then by the product rule of differentiation and some clever factoring

[tex] g'_n = (\prod^n_{k=1}f_{k})\sum^n_{j=1}\frac{f'_j}{f_j}[/tex]

which is even more simply written as

[tex] g'_{n} = g_{n}\sum^n_{j=1}\frac{f'_j}{f_j} [/tex].

To see how I get to this generalization I will work out the derivative for

[tex] g_{3} = \prod^3_{k=1}f_{k}=f_1f_2f_3 [/tex].

First,

[tex] g'_{3}= f_2f_3f'_1 + f_1(f_2f_3)'= f_2f_3f'_1 + f_1(f_2f'_3+f_3f'_2)= f_2f_3f'_1+f_1f_2f'_3+f_1f_3f'_2 [/tex]

Then noticing the relationship between this form and the original equation,

[tex] g'_3=\frac{g_3}{f_1}f'_1+\frac{g_3}{f_2}f'_2+\frac{g_3}{f_3}f'_3= g_3(\frac{f'_1}{f_1}+\frac{f'_2}{f_2}+\frac{f'_3}{f_3})[/tex]

which is what the general function predicted,

[tex] g'_3=g_3\sum^3_{j=1}\frac{f'_j}{f_j}[/tex]

I know that this statement is true for specific values of n, but I’m not sure what features this function has when g is an infinite product of differentiable functions. I hope that this helps with the question at hand.

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