How to differentiate series $\sum_{n=1}^{100} n(101 - n) \times \log(x - n)$

How to differentiate series $\sum\limits_{n =1}^{100} n(101 - n) \times \log(x - n)$?

I was solving a problem which is mentioned below:

If $f(x) = \prod\limits_{n=1}^{100} (x-n)^{n(101 - n)}$ then find $\dfrac{f(101)}{f'(101)}$.

I took log on both sides of the given function and I am unable to find the derivative of the series.

Can anyone just give me a hint how to differentiate the series as I'm just a beginner to this topic. I would be greatly thankful.

Kudos to the OP for being wary of assuming that $F(x)=\sum f_n(x)$ implies $F'(x)=\sum f_n'(x)$, but for finite sums that assumption is indeed justified. Consequently, the derivative of $F(x)=\sum_{n=1}^{100}n(101-n)\log(x-n)$ is

$$F'(x)=\sum_{n=1}^{100}{n(101-n)\over x-n}$$

and so

$$F'(101)=\sum_{n=1}^{100}n={100\cdot101\over2}=5050\quad\text{(Ligget se!)}^*$$

The justification for bringing the derivative inside a finite sum is essentially just mathematical induction on the linearity formula $(f+g)'=f'+g'$. It's only when you have an infinite sum that things get dicey and you have to be careful.

$^*$See Versions of the Gauss Schoolroom Anecdote collected by Brian Hayes.