If $x=log \, t $ and $y =\frac{1}{t^2}, $ then $\frac{d^2y}{dx^2}$ is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{4}{t^2}$
$x=\log t$
$\frac{dx}{dt}=\frac{1}{t}$
and,
$y=\frac{1}{t^2}$
$\frac{dy}{dt}=\frac{-2}{t^3}$
$⇒\frac{dy}{dx}=\frac{-2}{t^2}$
$⇒\frac{d^2y}{dx^2}=\frac{4}{t^3}×\frac{dt}{dx}=\frac{4}{t^2}$