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)?!
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1$\begingroup$ This follows from the usual product rule. Try writing it out for a cubic, you'll see the pattern. $\endgroup$ – lulu Jan 22 at 17:38
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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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1$\begingroup$ Ohh I get it now, I just never used the product rule for more than 2 factors. Thanks!!! $\endgroup$ – Happiness Is The Truth Jan 22 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.
