CUET Preparation Today
If $x=log_et, y=\frac{1}{t}$ then $\frac{d^2y}{dx^2}+\frac{dy}{dx}$ is equal to: |
0 $2e^{-x}$ $e^x$ $-2e^{-x}$ |
0 |
The correct answer is Option (1) → 0 $y=\frac{1}{t}$ and $x=\log t$ $⇒\frac{dy}{dt}=-\frac{1}{t^2}$ and $\frac{dx}{dt}=\frac{1}{t}$ $⇒\frac{dy}{dx}=-\frac{1}{t}⇒\frac{d^2y}{dx^2}=\frac{1}{t^2}×t=\frac{1}{t}$ $∴\frac{d^2y}{dx^2}+\frac{dy}{dx}=0$ |
