CUET Preparation Today
If $x=log \, t $ and $y =\frac{1}{t^2}, $ then $\frac{d^2y}{dx^2}$ is : |
$\frac{2}{t^2}$ $\frac{4}{t^2}$ $-\frac{1}{t}$ $-\frac{4}{t^2}$ |
$\frac{4}{t^2}$ |
The correct answer is Option (2) → $\frac{4}{t^2}$ $x=\log t$ $\frac{dx}{dt}=\frac{1}{t}$ and, $y=\frac{1}{t^2}$ $\frac{dy}{dt}=\frac{-2}{t^3}$ $⇒\frac{dy}{dx}=\frac{-2}{t^2}$ $⇒\frac{d^2y}{dx^2}=\frac{4}{t^3}×\frac{dt}{dx}=\frac{4}{t^2}$ |
