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How to take the derivative of $\prod_{i=1}^{n}(1-e^{-\lambda _{i}\cdot x})I(x>0)$ with regard to x?

Here $F(X)=\prod_{i=1}^{n}(1-e^{-\lambda _{i}\cdot x})I(x>0)$ is a CDF, and I want to take the derivative of it and get the pdf of X.

I try to take the log of F(X), so $\frac{\partial}{\partial x}logF(x)=\frac{\partial logF(x)}{\partial F(x)}\cdot \frac{\partial F(x)}{\partial x}$, then I can get $f(x)= \frac{\partial F(x)}{\partial x} $, but I don't know how the move on with $\frac{\partial}{\partial x}logF(x)$

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$$ \frac{d}{dx}\prod_{i=1}^{n}(1-e^{-\lambda _{i}x})I(x>0)=\frac{d}{dx}\left(\prod_{i=1}^{n}(1-e^{-\lambda _{i}x})\right)I(x>0)+\prod_{i=1}^{n}(1-e^{-\lambda _{i}x})\frac{d}{dx}I(x>0)\ . $$ Note that $$ \frac{d}{dx}I(x>0)=\delta(x)\ , $$ therefore in principle your pdf would have a Dirac mass at $x=0$, whose coefficient is $\prod_{i=1}^{n}(1-e^{-\lambda _{i}x})\Big|_{x=0}=0$ (so no point mass after all...).

Evaluating $\frac{d}{dx}\left(\prod_{i=1}^{n}(1-e^{-\lambda _{i}x})\right)$ is not hard, just rewrite $$ \frac{d}{dx}\left(\prod_{i=1}^{n}(1-e^{-\lambda _{i}x})\right)=\frac{d}{dx}e^{\sum_{i=1}^n \log(1-e^{-\lambda _{i}x})}=e^{\sum_{i=1}^n \log(1-e^{-\lambda _{i}x})}\sum_{i=1}^n \frac{d}{dx}\log(1-e^{-\lambda _{i}x}) $$ $$ =\left(\prod_{i=1}^{n}(1-e^{-\lambda _{i}x})\right)\sum_{j=1}^n\frac{\lambda_j e^{-\lambda_j x}}{1-e^{-\lambda_j x}}\ . $$

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