Question

lf $x=\log t$ and $y=t^2-1$, then $y^{\prime \prime }$ at $t=1$ is

• A. $2$
• B. $4$
• C. $3$
• D. $0$

Answer 1

Answer: (B)

$4$

$x=\log t\Rightarrow t=e^x$
At $t=1,x=0$.
Now, $y=t^2-1$
$y=e^{2x}-1$
$\Rightarrow y^{\prime \prime }=4e^{2x}$
$\Rightarrow y^{\prime \prime }=4$ at $x=0$
Option B is correct.