Question

If $x={\sin }^{-1}t$ and $y=\log (1-t^2)$, then $\frac{d^{2}y}{d{x}^2}$ at $t=1/2$ is

• A. $\frac{-8}{3}$
• B. $\frac{8}{3}$
• C. $\frac{3}{4}$
• D. $\frac{-3}{4}$

Answer 1

The correct option is A $\frac{-8}{3}$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{-2t}{1-t^2}}{\frac{1}{\sqrt{1-t^2}}}=\frac{-2t}{\sqrt{1-t^2}}$
$\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\ast \frac{dt}{dx}=\left(\frac{2\left(\sqrt{1-t^2}\right)-2t\left(\frac{t}{\sqrt{1-t^2}}\right)}{1-t^2}\right)(1-t^2)$
$=\left(\frac{-2(1-t^2)-2t^2}{1-t^2}\right)$
$\frac{d^{2}y}{d{x}^2}{|}_{t{=}^{1/2}}=\frac{-2}{3/4}$
$=\frac{-8}{3}$