# Find values of $f'(2)$ for given expression

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

• "When I put zero whole expression become zero" are you sure about this? what happens to $\frac{1}{x-2}$? Jul 5, 2016 at 17:25
• Sorry I meant to write 2 and this will become not defined Jul 5, 2016 at 17:26
• 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. Jul 5, 2016 at 17:34

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