Question

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

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

Answer 1

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

Given, $x={\sin }^{-1}t$ or $\sin x=t$

and $y=\log (1-t^2)$

$=\log (1-{\sin }^{2}x)$

$=\log \left({\cos }^{2}x\right)$

Hence, $y=2\log \left(cosx\right)$

Therefore, $\frac{dy}{dx}=\frac{-2\sin x}{\cos x}=-2\tan x$

$\frac{d^{2}y}{d{x}^2}=-2{\sec }^{2}x=\frac{-2}{{\cos }^{2}x}$

$=\frac{-2}{1-{\sin }^{2}x}=\frac{-2}{1-t^2}$

Hence, $\frac{d^{2}y}{d{x}^2}{|}_{t=\frac{1}{2}}=\frac{-2}{1-{\left(\frac{1}{2}\right)}^2}=\frac{-2}{1-\frac{1}{4}}=\frac{-8}{4-1}=\frac{-8}{3}$.