Explanation

(a)
Given that, $x = e^{t}\sin{t,\ y = e^{t}\cos{t\ldots(i)}}$
At point (1,1), $1 = e^{t}\sin{t,\ 1 = e^{t}\cos t}$
$\tan{t = 1\ \Rightarrow t = \frac{\pi}{4}}$
On differentiating Eq. (i) w.r.t. $x$, we get
$\frac{dy}{dt} = e^{t}(\cos{t - \sin{t)}}$
and $\frac{dx}{dt} = e^{t}(\sin{t + \cos{t)}}$
$\therefore\ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\cos{t - \sin t}}{\cos{t + \sin t}}$
Again differentiating w.r.t. $x$, we get
$\frac{d^{2}y}{dx^{2}} = \frac{d}{dt}\left( \frac{\cos{t - \sin t}}{\cos{t - \sin t}} \right)\frac{dt}{dx}$
$= \left\lbrack \frac{\begin{array}{r}
\lbrack(\cos{t + \sin{t)( - \sin{t - \cos{t) -}}}} \\
{(cos}{t - \sin{t)( - \sin{t + \cos{t)\rbrack}}}}
\end{array}}{(\cos{t + \sin{t\left. \ \ \right)^{2}\ }}} \right\rbrack\frac{dt}{dx}$
$= \frac{- 2}{\left( \cos{t + \sin t} \right)^{2}}.\frac{1}{e^{t}(\sin{t + \cos{t)}}}$
$= \ \ \frac{- 2}{(e^{t}\cos{t + e^{t}\sin{t)}}}.\frac{1}{\left( \cos{t + \sin t} \right)^{2}}$
$= \frac{- 2}{x + y}.\frac{1}{(\cos{t + \sin{t\left. \ \ \right)^{2}\ }}}\ \ \lbrack from\ Eq.(i)\rbrack$
At $t = \frac{\pi}{4},x = 1,\ y = 1$
$\therefore\frac{d^{2}y}{dx^{2}} = \frac{- 2}{1 + 1}.\frac{1}{\left( \cos\frac{\pi}{4} + \sin\frac{\pi}{4} \right)^{2}}$
$= \frac{- 1}{\left\lbrack \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right\rbrack} = - \frac{1}{2}$