Explanation

(c)
Let $y = a\sin^{3}{t,\ \ x = a\cos^{3}t}$
On differentiating w.r.t. $t$, we get
$\frac{dy}{dt} = 3a\sin^{2}{t\cos{t,\ \frac{dx}{dt} = - 3a\cos^{2}{t\sin t}}}$
$\therefore\ \ \frac{dy}{dx} = \frac{3{a\ \sin}^{2}{t\cos t}}{- 3a\cos^{2}{t\sin t}}\ \$
$\Rightarrow \ \frac{dy}{dx} = \ - \frac{\sin t}{\cot t} = - \tan t$
Again, differentiating w.r.t. $x$, we get
$\frac{d^{2}y}{dx^{2}} = \ - \sec^{2}{t.\frac{dt}{dx}}$
$= - \frac{\sec^{2}t}{- 3a\cos^{2}{t\ s}in\ t}$
$= \frac{1}{3a\cos^{4}{t\sin t}}$
$\Rightarrow \ \ \left( \frac{d^{2}y}{dx^{2}} \right)_{t = \frac{\pi}{4}} = \frac{1}{3a\left( \frac{1}{\sqrt{2}} \right)^{4}.\left( \frac{1}{\sqrt{2}} \right)}$
$= \frac{\left( \sqrt{2} \right)^{5}}{3a} = \frac{4\sqrt{2}}{3a}$