# What is the derivative of an iterated product like $\frac{d}{dx}\prod \limits_{i=0}^n \ln(y_i^{x - 1})$?

Suppose the following function with pi notation, with the pi denoting the iterated product, multiplying from $$i = 0$$ to $$i = n$$:

$$\prod_{i=0}^n \ln(y_i^{x - 1})$$

That is, the natural logarithm of $$y$$, subscripted by $$i$$, to the power of $$x - 1$$.

What is the derivative of this product - to be clear, its derivative with respect to $$x$$, not $$y$$?

• A few further comments, now on the problem itself. What are the $y_i$'s, necessarily? Also, have you considered simply brute forcing with the product rule? There's a generalization for when you have more than $2$ functions as a product and you're trying to take the derivative of them. It's touched on in this MSE post -- math.stackexchange.com/questions/1348251/… – Eevee Trainer Mar 1 '19 at 2:30
• Hi sorry this is my first post on MSE so I didn't know to use MathJax. Thank you for editing. I am not able to solve this as I don't understand how to proceed, and searching for the answer has proved fruitless thus I posted here. – user8115948 Mar 1 '19 at 2:35

We don't even need a product rule. We have \begin{align*} g(x) = \prod_{i=0}^{n}\ln(y_i^{x-1}) = (x-1)^{n+1}\prod_{i=1}^{n}\ln(y_i) \end{align*} And so \begin{align*} \frac{d}{dx} g(x) = (n+1)(x-1)^{n}\prod_{i=0}^{n}\ln(y_i) \end{align*}
Define $$\alpha=\prod_{i=0}^n \ln(y_i)$$. Note that $$\alpha$$ is simply a constant. Use logarithm properties, use the power rule for derivatives and you're done.
$$f(x)=\prod_{i=0}^{n}\ln{y_i}^{x-1}=\prod_{i=0}^{n}(x-1)\ln{y_i}$$$$\alpha(x-1)^{n+1}\implies \dfrac{\mathrm df}{\mathrm dx}=\alpha(n+1)(x-1)^n$$ $$\boxed{\dfrac{\mathrm d}{\mathrm dx}\prod_{k=0}^n \ln{y_i}^{x-1}=(n+1)(x-1)^n\prod_{k=0}^n\ln y_i}$$
Let $$f(x)=\prod_{i=0}^{n}f_i(x)$$ and let $$g^{(m)}(x)=\left(\frac{d}{dx}\right)^mg(x),\qquad m=0,1,2,...$$ as well as $$\delta_{ij}$$ denote the Kronecker Delta.
We have that $$f'(x)=\sum_{i=0}^{n}\prod_{j=0}^{n}f_j^{(\delta_{ij})}(x)=\sum_{i=0}^{n}\frac{f_i'(x)}{f_i(x)}f(x)$$ We use this with the choice $$f_i(x)=\ln(y_i^{x-1})=(x-1)\ln y_i$$ which gives $$f_i'(x)=\ln y_i$$ So $$f'(x)=\sum_{i=0}^{n}\frac{1}{x-1}\prod_{j=0}^{n}(x-1)\ln y_j$$ $$f'(x)=(x-1)^n\sum_{i=0}^{n}\prod_{j=0}^{n}\ln y_j$$ $$f'(x)=(x-1)^n\left(\sum_{i=0}^{n}1\right)\prod_{j=0}^{n}\ln y_j$$ $$f'(x)=(n+1)(x-1)^n\prod_{j=0}^{n}\ln y_j$$