Explanation

(d)
Since, $x = e^{t}\sin{t\ \ and\ \ y = e^{t}\cos t}$
$\Rightarrow \ \ \ \ \frac{dx}{dt} = e^{t}\cos{t + \sin{t\ e^{t}}}$
and $\frac{dy}{dt} = \ - e^{t}\sin{t + e^{t}\cos t}$
$\therefore\ \ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{(\cos{t - \sin{t)}}}{(\cos{t + \sin{t)}}}$
$\frac{d^{2}y}{dx^{2}} = \ \frac{\begin{bmatrix}
(\cos{t + \sin{t)( - \sin{t - \cos{t)}}}} \\
- (\cos{t - \sin{t)( - \sin{t + \cos{t)}}}}
\end{bmatrix}}{{(\cos{t + \sin{t)}}}^{2}}$ $\frac{dt}{dx}$
$= \frac{- {(\sin{t + \cos{t)}}}^{2} - {(\cos{t - \sin{t)}}}^{2}}{{(\cos{t + \sin{t)}}}^{2}}$
$\times \frac{1}{e^{t}(\cos{t + \sin{t)}}}$
$= - \frac{2}{e^{t}{(\cos{t + \sin{t)}}}^{3}}$
$\Rightarrow \left( \frac{d^{2}y}{dx^{2}} \right)_{(x = \pi)} = \frac{- 2}{e^{\pi}{(\cos{\pi + \sin{\pi)}}}^{3}} = \frac{2}{e^{\pi}}\ \$