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Let $f(x) = (x-2)(x-4)(x-6).........(x-2n)$ then $f'(2)$ equals

My work $$f'(x)=(x-4)(x-6)...(x-2n) + (x-2)(x-6)...(x-2n) + (x-2)(x-4)...(x-2n)... (x-2)(x-4)(x-6)...(x-2n -2)$$ $$f'(x)=(x-2)(x-4)(x-6)...(x-2n)[\frac{1}{(x-2)}+\frac{1}{(x-4)} +\frac{1}{(x-6)} +...\frac{1}{(x-2n)}] $$ When I put two whole expression will be not defined but this is not the answer. Please tell me how to solve this

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  • $\begingroup$ "When I put zero whole expression become zero" are you sure about this? what happens to $\frac{1}{x-2}$? $\endgroup$
    – Nikunj
    Jul 5, 2016 at 17:25
  • $\begingroup$ Sorry I meant to write 2 and this will become not defined $\endgroup$
    – Aakash Kumar
    Jul 5, 2016 at 17:26
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    $\begingroup$ You can't factor $(x-2)$ out of the entire expression and still get the answer. You can factor it out of most of the expression, as hinted at in the answer below. $\endgroup$
    – Cameron Buie
    Jul 5, 2016 at 17:34

1 Answer 1

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Hint

Write $f(x)=(x-2)g(x)$

Then $f'(x)=g(x)+g'(x)(x-2)$.

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  • $\begingroup$ Why I am getting wrong answer with my method $\endgroup$
    – Aakash Kumar
    Jul 5, 2016 at 17:29
  • $\begingroup$ @AakashKumar The two expressions that you’ve written down for $f'(x)$ in your question are not equivalent. The latter is undefined at the even positive integers $\le 2n$, which you yourself have noted for $x=2$, while the former is defined everywhere. $\endgroup$
    – amd
    Jul 5, 2016 at 17:36

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