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P(x) is a polynomial with roots x1,x2,x3...xn and P'(x) is its derivative. So can anyone explain how P'(x) can be factored as shown in the screenshot, using the roots of P(x)?!enter image description here

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    $\begingroup$ This follows from the usual product rule. Try writing it out for a cubic, you'll see the pattern. $\endgroup$
    – lulu
    Jan 22, 2020 at 17:38

3 Answers 3

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Let $f_k(x) = x-x_k$, note that $f_k'(x) = 1$.

Since $P(x) = f_1(x) f_2(x)\cdots f_n(x)$, the product differentiation rule gives $P'(x) = f_1'(x) f_2(x)\cdots f_n(x) + f_1(x) f_2'(x)\cdots f_n(x) + \cdots + f_1(x) f_2(x)\cdots f_n'(x)$, or $P'(x) = f_2(x)\cdots f_n(x) + f_1(x) f_3(x)\cdots f_n(x) + \cdots + f_1(x) f_2(x)\cdots f_{n-1}(x)$.

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    $\begingroup$ Ohh I get it now, I just never used the product rule for more than 2 factors. Thanks!!! $\endgroup$ Jan 22, 2020 at 18:31
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$P(x)=\prod (x-x_i)$ and so $\ln P(x)=\sum \ln (x-x_i)$

Differentiate w.r.t. $x$

$$\frac{P'(x)}{P(x)}=\sum \frac{1}{x-x_i}$$

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So you want to prove $${P'(x)\over P(x) } = {1\over x-x_1}+{1\over x-x_2}+...+{1\over x-x_n}$$

but this is the same as $$(\ln P(x))' = (\ln(x-x_1)+\ln(x-x_2)+...+\ln(x-x_n))'$$

or $$(\ln P(x))' = (\ln(x-x_1)(x-x_2)...(x-x_n))'$$ or $$\ln P(x) = \ln(x-x_1)(x-x_2)...(x-x_n)+const$$

If we write $a= e^{const}$ i.e. $const = \ln a$ we get $$p(x) =a(x-x_1)(x-x_2)...(x-x_n)$$

Now read this from down to up and thus a conslusion.

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