Explanation

(b)
$\because\ f(x) = (x - 2)(x - 4)(x - 6)\ldots(x - 2n)$
Taking log on both sides in the given equation, we get
log $f(x) = \log{(x - 2) + \log{(x - 4) + ... + \log(x - 2n)}}$
on differentiating w.r.t. $x,\$we get
$\frac{1}{f(x)}f^{'}(x) = \frac{1}{(x - 2)} + \frac{1}{(x - 4)} + ... + \frac{1}{(x - 2n)}$
$\Rightarrow \ \ \ \ f^{'}(x) = (x - 4)(x - 6)\ldots(x - 2n)$
$+ (x - 2)(x - 6)\ldots(x - 2n)$
$+ \ldots + (x - 2)(x - 6)\ldots(x - 2(n - 1))$
$\therefore\ f^{'}(2) = ( - 2)( - 4)\ldots(2 - 2n)$
$= ( - 2)^{n - 1}(1.2\ldots..(n - 1) = ( - 2)^{n - 1}(n - 1)!$