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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 $y$?

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  • $\begingroup$ For future reference, it is preferred you write your posts in MathJax on MSE. I have already done that for you, with some rewording and such to clarify your intent. Generally, though, it will be preferred if you include your own attempts at this problem in your post - people on MSE respond better when you include your own attempts and work, your own understanding of the problem, and the context where this pops up. As-is, I feel your post will likely accumulate downvotes and close votes despite my edits. You're more than welcome to edit your post to include these details though. $\endgroup$ Mar 1, 2019 at 2:28
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    $\begingroup$ 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/… $\endgroup$ Mar 1, 2019 at 2:30
  • $\begingroup$ 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. $\endgroup$ Mar 1, 2019 at 2:35

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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*}

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  • $\begingroup$ Very well done! +1 $\endgroup$
    – clathratus
    Mar 1, 2019 at 5:07
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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.

$$f(x)=\prod_{i=0}^{n}\ln{y_i}^{x-1}=\prod_{i=0}^{n}(x-1)\ln{y_i}$$$$\alpha(x-1)^{n+1}\implies \dfrac{\mathrm df}{\mathrm dx}=\alpha(n+1)(x-1)^n$$ $$\boxed{\dfrac{\mathrm d}{\mathrm dx}\prod_{k=0}^n \ln{y_i}^{x-1}=(n+1)(x-1)^n\prod_{k=0}^n\ln y_i}$$

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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 $$f'(x)=\sum_{i=0}^{n}\frac{1}{x-1}\prod_{j=0}^{n}(x-1)\ln y_j$$ $$f'(x)=(x-1)^n\sum_{i=0}^{n}\prod_{j=0}^{n}\ln y_j$$ $$f'(x)=(x-1)^n\left(\sum_{i=0}^{n}1\right)\prod_{j=0}^{n}\ln y_j$$ $$f'(x)=(n+1)(x-1)^n\prod_{j=0}^{n}\ln y_j$$

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