Explanation

(c)
Given, $x = \log{\left( 1 + t^{2} \right)\ \ \ and\ \ y = t - \tan^{- 1}t}$
$\frac{dx}{dt} = \frac{1}{1 + t^{2}}.2t$
and $\frac{dy}{dt} = 1 - \frac{1}{1 + t^{2}} = \frac{t^{2}}{1 + t^{2}}$
$\therefore$ $\ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{t^{2}/(1 + t^{2})}{2t/(1 + t^{2})} = \frac{t}{2}$ …(i)
Also, $x = \log{\left( 1 + t^{2} \right) \Rightarrow t^{2} = e^{x} - 1}$ …(ii)
From Eqs. (i) and (ii), we get
$\frac{dy}{dx} = \frac{\sqrt{e^{x} - 1}}{2}$