Question

If $x=\log t$ and $y=t^2-1,$ then $y^{\prime \prime }\left(1\right)$ at $t=1$ is

• A. $2$
• B. $4$
• C. $3$
• D. None of these

Answer 1

Answer: (B)

$4$

$x=\log t\Rightarrow \frac{dx}{dt}=\frac{1}{t}$
$y=t^2-1\Rightarrow \frac{dy}{dt}=2t$
$\frac{dy}{dx}=y^{\prime }=2t^2$
$\frac{d{y}^{\prime }}{dx}=\frac{d{y}^{\prime }}{dt}\cdot \frac{dt}{dx}$
$\frac{d^{2}y}{d{x}^2}=\frac{4t}{\frac{1}{t}}=4t^2$
$\therefore y^{\prime \prime }\left(1\right)=4\left(1\right)=4$