# How to take the derivative of product of a sequence?

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 $$\frac{\partial}{\partial x}logF(x)$$

$$\frac{d}{dx}\prod_{i=1}^{n}(1-e^{-\lambda _{i}x})I(x>0)=\frac{d}{dx}\left(\prod_{i=1}^{n}(1-e^{-\lambda _{i}x})\right)I(x>0)+\prod_{i=1}^{n}(1-e^{-\lambda _{i}x})\frac{d}{dx}I(x>0)\ .$$ 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 $$\frac{d}{dx}\left(\prod_{i=1}^{n}(1-e^{-\lambda _{i}x})\right)=\frac{d}{dx}e^{\sum_{i=1}^n \log(1-e^{-\lambda _{i}x})}=e^{\sum_{i=1}^n \log(1-e^{-\lambda _{i}x})}\sum_{i=1}^n \frac{d}{dx}\log(1-e^{-\lambda _{i}x})$$ $$=\left(\prod_{i=1}^{n}(1-e^{-\lambda _{i}x})\right)\sum_{j=1}^n\frac{\lambda_j e^{-\lambda_j x}}{1-e^{-\lambda_j x}}\ .$$