Answer:
Option A
Explanation:
Given , x=f(t) and y=g(t)
On digfferentiating both sides w.r.t 't' , we get
$\frac{dx}{dt}=f'(t) $and $ \frac{dy}{dt}=g'(t)$
We know that, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\Rightarrow$ $\frac{dy}{dx}=\frac{g'(t)}{f'(t)}$
Again, differentiating both sides w.r.t 'x' , we get
$\frac{d^{2}y}{dx^{2}}=\frac{f'(t).g''(t)-g'(t).f''(t)}{(f'(t))^{2}}.\frac{dt}{dx}$
=$\frac{f'(t).g''(t)-g'(t).f''(t)}{(f'(t))^{2}}.\frac{1}{f'(t)}$
$ =\frac{f'(t).g''(t)-g'(t).f''(t)}{(f'(t))^{3}}$
