find a derivative

[Infernum]

Find the domain of existense and the derivative of the expression

[tex]f(x)=\prod_{k=1}^{n}(x-k)^x[/tex]

[shyamjayakannan]

To find the domain, we need to ensure that the function never takes the [tex]0^0[/tex] form. For this to happen the [tex]x[/tex] in the exponent would have to be 0, bt then the base is not 0. Which means this will never happen and so, the domain is [tex]x \in \mathbb{R}[/tex].

Now, [tex]f(x)=\prod_{k=1}^n(x-k)^x=\prod_{k=1}^ne^{x\ln(x-k)}=e^{x\sum_{k=1}^n\ln(x-k)}[/tex]

So, [tex]\frac{d}{dx}f(x)=\frac{d}{dx}\left\{e^{x\sum_{k=1}^n\ln(x-k)}\right\}=e^{x\sum_{k=1}^n\ln(x-k)}\frac{d}{dx}\left\{x\sum_{k=1}^n\ln(x-k)\right\}=e^{x\sum_{k=1}^n\ln(x-k)}\left\{\sum_{k=1}^n\ln(x-k)+x\frac{d}{dx}\sum_{k=1}^n\ln(x-k)\right\}...(1)[/tex]

Now, [tex]\frac{d}{dx}\sum_{k=1}^n\ln(x-k)=\sum_{k=1}^n\frac{d}{dx}\ln(x-k)=\sum_{k=1}^n\frac{1}{x-k}...(2)[/tex]

Substituting (2) into (1), we get: [tex]\boxed{\frac{d}{dx}f(x)=e^{x\sum_{k=1}^n\ln(x-k)}\left\{\sum_{k=1}^n\ln(x-k)+x\sum_{k=1}^n\frac{1}{x-k}\right\}}[/tex]