Explanation

(c)
Given, $y - 1 = t^{10},\ \ x - 1 = t^{8}$
$\Rightarrow \ \frac{dy}{dt} = 10t^{9}\ \ and\ \ \frac{dx}{dt} = 8t^{7}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$
$\therefore\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{5}{4}t^{2}$
$= \frac{5}{4}\frac{t^{10}}{t^{8}} = \frac{5(y - 1)}{4(x - 1)}$
$\Rightarrow \ \ \frac{d^{2}y}{dx^{2}} = \frac{5}{4}\ \frac{(x - 1)\frac{dy}{dx} - (y - 1)1\ }{(x - 1)^{2}}$
$= \frac{5}{4}\frac{1}{(x - 1)}\left\lbrack \frac{dy}{dx} - \frac{(y - 1)}{(x - 1)} \right\rbrack$
$= \frac{5}{4}\frac{1}{(x - 1)}\left\lbrack \frac{5}{4}.\frac{(y - 1)}{(x - 1)} - \frac{(y - 1)}{(x - 1)} \right\rbrack$
$= \frac{5}{4}\frac{(y - 1)}{(x - 1)^{2}}\left( \frac{5}{4} - 1 \right) = \frac{5}{16t^{6}}$