# Differentiate One Infinite Product: Strategies & Formulas

• the dude man
In summary, the conversation discusses differentiating functions and the derivative of an infinite product. The general formula for differentiating an infinite product is g'_n = g_n * (sum of f'_j/f_j), where g_n is the product of n differentiable functions. A proof is provided for the specific case of n = 3. The conversation also addresses the difference between an infinite product and an infinite sum.
the dude man
How do you differentiate one? Is it possible? Any formulas?

Apply d/dx to the expression.

One differentiates functions, not numbers. Do you mean you want to know the derivative of the identity function? Over what domain? Or do you want to know the derivative of the function that always returns 1?

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

M=
----
\ f' / f
/
----
Does that work?

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

I can't figure this out can someone refer me to proof cause i can't find one.

The derivative of a function in the form of an infinite product.
Anything would be nice.

Bump. .

Please don't bump. You asked how to differentiate an infinite product, then posted an ascii art diagram that indicates an infinite sum, not product. Which is it? If you take time to learn how to post latex here it will help you.

equation for derivative of n product

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 $$g_{n} = \prod^n_{k=1}f_{k}$$ and assume that $$f'_n$$ exists. Then by the product rule of differentiation and some clever factoring
$$g'_n = (\prod^n_{k=1}f_{k})\sum^n_{j=1}\frac{f'_j}{f_j}$$
which is even more simply written as
$$g'_{n} = g_{n}\sum^n_{j=1}\frac{f'_j}{f_j}$$.

To see how I get to this generalization I will work out the derivative for
$$g_{3} = \prod^3_{k=1}f_{k}=f_1f_2f_3$$.
First,
$$g'_{3}= f_2f_3f'_1 + f_1(f_2f_3)'= f_2f_3f'_1 + f_1(f_2f'_3+f_3f'_2)= f_2f_3f'_1+f_1f_2f'_3+f_1f_3f'_2$$
Then noticing the relationship between this form and the original equation,
$$g'_3=\frac{g_3}{f_1}f'_1+\frac{g_3}{f_2}f'_2+\frac{g_3}{f_3}f'_3= g_3(\frac{f'_1}{f_1}+\frac{f'_2}{f_2}+\frac{f'_3}{f_3})$$
which is what the general function predicted,
$$g'_3=g_3\sum^3_{j=1}\frac{f'_j}{f_j}$$

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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## 1. What is the purpose of differentiating one infinite product?

The purpose of differentiating one infinite product is to find its rate of change, or how quickly the product is changing with respect to the variables involved. This can help with solving optimization problems and understanding the behavior of the product.

## 2. What are some strategies for differentiating one infinite product?

One strategy is to use the product rule, which states that the derivative of a product of two functions is equal to the first function times the derivative of the second function, plus the second function times the derivative of the first function. Another strategy is to use logarithmic differentiation, which can be helpful for products with large exponents or multiple variables.

## 3. Can you provide an example of differentiating one infinite product?

Sure, let's say we have the infinite product f(x) = x^2 * (1/2)^x. Using the product rule, we would take the derivative of x^2 (2x) and multiply it by the second function (1/2)^x, then add the derivative of (1/2)^x (-ln(2) * (1/2)^x) multiplied by the first function (x^2). This gives us f'(x) = 2x * (1/2)^x + x^2 * (-ln(2) * (1/2)^x).

## 4. Are there any special formulas for differentiating one infinite product?

Yes, there is a special formula for differentiating infinite products of the form (1+x)^n, where n is a constant. The derivative can be calculated using the generalized binomial theorem, which states that (1+x)^n = ∑(n choose k)*x^k, where k ranges from 0 to n, and (n choose k) = n!/(k!(n-k)!). The derivative would then be f'(x) = n*(1+x)^(n-1).

## 5. How can differentiating one infinite product be applied in real-world situations?

One example is in finance and economics, where infinite products are used in time value of money calculations. By differentiating these products, we can determine the rate of change of the present value or future value of money with respect to interest rates or time. This can be useful in making investment decisions and understanding economic trends.

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