What is $g'(x)$ at $c_j$

Question

$F$ is a field and $c_1,...c_n$ are different scalars in $F$.

$$g(x)=\prod_{i=1}^{n}(x-c_i)$$

What is $g'(c_j)$?

I looked here, I don't understand gow can I put there $c_j$...

If I calculated it right: $$g'\left(x\right)=\left(\prod_{i=1}^{n}\left(x-c_{i}\right)\right)\cdot\left(\sum_{i=1}^{n}\frac{1}{\left(x-c_{i}\right)}\right)$$ and I can't put there $c_j$ because I dividing by zero...

Can you help me please?

Thank you!

Answer 1

You don't really need to divide. See what happens when $n=2$

The derivative of $(x-c_1)(x-c_2)$ becomes $(x-c_1)+(x-c_2)$ (no division)

You wrote $g'(x)$ as a product of two terms, however there are cancellations. Once you remove the cancellations you have

$$g'(x) = \sum_{i=1}^n \prod_{k\not = i} (x-c_k)$$

Now insert $x=c_j$, then $\prod_{k\not = i} (c_j-c_k)=0$ unless $i=j$. In this case we get $\prod_{k\not=j} (c_j-c_k)$.

Answer 2

The formula you have derived is valid at all points $x$ other than the $c_j$'s. $g'(c_j)$ is simply $\prod_{i \neq j}(c_j-c_i)$