Let $f(x) = \frac{x(x−1)(x−2)\cdots(x−n)}{(x + 1)(x + 2)\cdots(x + n)}$ It's my first time asking questions on this community. Could you guys help me? 
Let $$f(x) = \frac{x(x−1)(x−2)\cdots(x−n)}{(x + 1)(x + 2)\cdots(x + n)}$$
a) Find $f '(0)$
b) Suppose that the function $f: \mathbb{R}\to\mathbb{R}$ satisfies $f(x+y) = f(x)f(y)$ for all $x, y \in\mathbb{R}$. 
If $k:=f '(0)$, show that $f '(x)= k f(x)$.
(Consider the cases $f (0) = 0$ and $f(0)$ is not the same as $0$ separately.)
Thank you so much
( I've been attempting to solve the problem (A) by multiplying and dividing the function in order to simplify the function and looking for the "logic" of this problem and yet I still don't know how to solve the problem. Sorry if you guys feel that way! I'll try to solve it again, I'm so sorry :) )
Update: 
For problem A, I've tried and found the answer by using an internet calculator (wolfram alpha) So, here it is:
if n is an odd number, the answer will be=-1 
if n is an even number, the answer will be=1.
But I am not too sure about it and therefore, I need strong proof that can prove my answer and I really hope that you guys could help me.
 A: For part a let us go to the limit definition of a derivative. That is, the derivative of a function at point a is:
$$\frac{df}{dx_a}=\lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}$$
Consequently, the function becomes:
$$\lim\limits_{h\to 0}\frac{\frac{0+h(0+h−1)(0+h−2)\cdots(0+h−n)}{(0+h + 1)(0 +h+ 2)\cdots(0+h + n)}-\frac{0(0−1)(0−2)\cdots(0−n)}{(0 + 1)(0 + 2)\cdots(0 + n)}}{h}$$
$$\lim\limits_{h\to 0}\frac{\frac{h(h−1)(h−2)\cdots(h−n)}{(h + 1)(h+ 2)\cdots(h + n)}-\frac{0}{(1)( 2)\cdots( n)}}{h}$$
$$\lim\limits_{h\to 0}\frac{\frac{h(h−1)(h−2)\cdots(h−n)}{(h + 1)(h+ 2)\cdots(h + n)}-0}{h}$$
$$\lim\limits_{h\to 0}\frac{\frac{h(h−1)(h−2)\cdots(h−n)}{(h + 1)(h+ 2)\cdots(h + n)}}{h}$$ 
After simplifying the rational we get:
$$\lim\limits_{h\to 0}\frac{h(h−1)(h−2)\cdots(h−n)}{h(h + 1)(h+ 2)\cdots(h + n)}$$
Cancelling out $h$, we get: 
$$\lim\limits_{h\to 0}\frac{(h−1)(h−2)\cdots(h−n)}{(h + 1)(h+ 2)\cdots(h + n)}$$
After letting the limit go to zero:
$$\lim\limits_{h\to 0}\frac{(h−1)(h−2)\cdots(h−n)}{(h + 1)(h+ 2)\cdots(h + n)}=\frac{(-1)(-2)\cdots (-n)}{(1)(2)\cdots (n)}=-1$$
Therefore, the derivative at $0$ is -1.
Part b is very similar to part one, you just have to use the limit definition of derivative, and then plug-in the disjunctive values given.
