Explanation

(c)
$y = \cos^{- 1}\left( \frac{1 - \log x}{1 + \log x} \right)$
On differentiating w.r.t. $x,\$we get
$\frac{dy}{dx} = \ - \frac{1}{\sqrt{1 - \left( \frac{1 - \log x}{1 + \log x} \right)^{2}}}$
$\frac{(1 + \log{x)\left( \frac{- 1}{x} \right) - (1 - \log{x)\left( \frac{1}{x} \right)}}}{{(1 + \log{x)}}^{2}}$
$= - \frac{1}{2\sqrt{\log x}}.\frac{- 1 - \log{x - 1 + \log x}}{x(1 + \log{x)}}$
$= \frac{1}{x\sqrt{\log x}(1 + \log{x)}}$
$\therefore\left( \frac{dy}{dx} \right)_{x = e} = \frac{1}{e(1 + 1)} = \frac{1}{2e}$