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 $\frac{d}{d x}logF(x)=\frac{d logF(x)}{d F(x)}\cdot \frac{d F(x)}{d x}$,
@Ankit Kumar help me with:
$$\frac{d(logF(x))}{dx}=\sum_{i=1}^n\frac{d(log(1-e^{-\lambda_ix}))}{dx}$$, then $$\frac{1}{F(x)}\frac{d(F(x))}{dx}=\sum_{i=1}^n\frac{1}{\log(1-e^{-\lambda_ix})}{\lambda_ie^{-\lambda_ix}}$$, but I don't know how to move on with $\frac{d(F(x))}{dx}=\sum_{i=1}^n\frac{1}{\log(1-e^{-\lambda_ix})}{\lambda_ie^{-\lambda_ix}}\cdot \prod_{i=1}^{n}(1-e^{-\lambda _{i}\cdot x})$, how to simplify it? since it is a multiplication of a sum of sequence and a product of sequence.
Here is another approach. Recall the product formula $(f\cdot g)^{\prime}=f^{\prime}\cdot g+f\cdot g^{\prime}$ of the product of two differentiable functions $f,g$. We can generalise this (and for instance prove it with induction) and obtain \begin{align*} \color{blue}{\left(\prod_{j=1}^n f_j(x)\right)^{\prime}} &=\left(f_1(x)\right)^\prime \prod_{j=2}^nf_j(x)+f_1(x)\left(\prod_{j=2}^n f_j(x)\right)^\prime\\ &=\left(f_1(x)\right)^\prime \prod_{j=2}^nf_j(x)+f_1(x)\left(\left(f_2(x)\right)^\prime \prod_{j=3}^nf_j(x)+f_2(x)\left(\prod_{j=3}^n f_j(x)\right)^\prime\right)\\ &=\left(f_1(x)\right)^\prime \prod_{{j=1}\atop{j\ne 1}}^nf_j(x)+\left(f_2(x)\right)^\prime \prod_{{j=1}\atop{j\ne 2}}^nf_j(x)+f_1(x)f_2(x)\left(\prod_{j=3}^n f_j(x)\right)^\prime\\ &\ \ \vdots\\ &=\left(f_1(x)\right)^\prime \prod_{{j=1}\atop{j\ne 1}}^nf_j(x)+\left(f_2(x)\right)^\prime \prod_{{j=1}\atop{j\ne 2}}^nf_j(x) +\cdots+\left(f_n(x)\right)^\prime \prod_{{j=1}\atop{j\ne n}}^nf_j(x)\\ &\,\,\color{blue}{=\sum_{k=1}^n\left(f_k(x)\right)^{\prime} \prod_{{j=1}\atop{j\ne k}}^nf_j(x)}\tag{1} \end{align*}