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