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