Explanation

(b)
Given, $y = e^{x}.e^{x^{2}}.e^{x^{3}}\ldots.e^{x^{n}}\ldots.$
$\Rightarrow \ \ \ \ \ \ \ \ y = e^{(x + x^{2} + \ldots + \infty)}$
$\Rightarrow \ \ \ \ \ \ \ \ y = e^{\frac{x}{1 - x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$
$\Rightarrow {\ \ \ \ \ \ \frac{dy}{dx} = e}^{\frac{x}{1 - x}}\left\lbrack \frac{(1 - x)1 - x( - 1)}{\left( 1 - x^{2} \right)} \right\rbrack = e^{\frac{x}{1 - x}}.\frac{1}{(1 - x)^{2}}$
At $x = \frac{1}{2},$
$\left( \frac{dy}{dx} \right)_{x = \frac{1}{2}} = e^{\frac{1/2}{1 - 1/2}}.\ \frac{1}{\left( 1 - \frac{1}{2} \right)^{2}} = 4e$