If $x=\frac{1}{t^2}, y =\frac{1}{t}$, then value of $\frac{d^2y}{dx^2}$ at $t=2$ is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → -2
$x=\frac{1}{t^2}, y =\frac{1}{t}$
$⇒\frac{dx}{dt}=\frac{-2}{t^3},⇒\frac{dy}{dt}=\frac{-1}{t^2}$
$\frac{dy}{dx}=\frac{-1}{t^2}×\frac{t^3}{-2}=\frac{t}{2}$
$⇒\frac{d^2y}{dx^2}=\frac{1}{2}×\frac{t^3}{-2}=\frac{t^3}{-4}$
$⇒\left.\frac{d^2y}{dx^2}\right|_{t=2}=\frac{8}{-4}=-2$