If $x=log_et, y=\frac{1}{t}$ then $\frac{d^2y}{dx^2}+\frac{dy}{dx}$ is equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
$y=\frac{1}{t}$ and $x=\log t$
$⇒\frac{dy}{dt}=-\frac{1}{t^2}$ and $\frac{dx}{dt}=\frac{1}{t}$
$⇒\frac{dy}{dx}=-\frac{1}{t}⇒\frac{d^2y}{dx^2}=\frac{1}{t^2}×t=\frac{1}{t}$
$∴\frac{d^2y}{dx^2}+\frac{dy}{dx}=0$