MHB Derivatives of symmetric expressions

[kalish1]

So I was bored in math class and came up with this series of related questions, that I cannot answer:

Is there a clean expression for $f'(x),$ where $$f(x)=\prod_{i=1}^{n}\dfrac{(x-i)}{(x+i)}?$$

What about for $f''(x)?$ Or for $$f(x)=\prod_{i=1}^{n}\dfrac{(x^2-i)}{(x^2+i)}?$$

 

[chisigma]

So I was bored in math class and came up with this series of related questions, that I cannot answer:

Is there a clean expression for $f'(x),$ where $$f(x)=\prod_{i=1}^{n}\dfrac{(x-i)}{(x+i)}?$$

What about for $f''(x)?$ Or for $$f(x)=\prod_{i=1}^{n}\dfrac{(x^2-i)}{(x^2+i)}?$$

You can use the relation...

$\displaystyle \frac{d}{d x} \ln f(x) = \frac{f^{\ '} (x)}{f(x)} \implies f^{\ '} (x) = f(x)\ \frac{d}{d x} \ln f(x)\ (1)$

... and in this case...

$\displaystyle \frac{d}{d x} \ln f(x) = \sum_{i = 1}^{n} (\frac{1}{x - i} - \frac{1}{x + i}) = 2\ \sum_{i=1}^{n} \frac{i}{x^{2} - i^{2}}\ (2)$

The same procedure is valid also for the second function...

Kind regards

$\chi$ $\sigma$