Question

If $x={\text{sin}}^{-1}t$ and $y=\text{log}(1-t^2)$; then ${\begin{array}{c}\frac{d^{2}y}{d{x}^2}\end{array}|}_{t=\frac{1}{2}}$ is

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

Answer 1

The correct option is D

$-\frac{8}{3}$

$\frac{dx}{dt}=\frac{1}{\sqrt{1-t^2}},\frac{dy}{dt}=-\frac{2t}{1-t^2}\Rightarrow \frac{dy}{dx}=\frac{-2t}{\sqrt{1-t^2}}$

$\frac{d^{2}y}{d{x}^2}=-2\frac{d}{dt}\left(\frac{t}{\sqrt{1-t^2}}\right)\frac{dt}{dx}$

$=-2\frac{(\sqrt{1-t^2}+\frac{t^2}{\sqrt{1-t^2}})\times \frac{dt}{dx}}{(1-t^2)}=-2\frac{1}{1-t^2}$

$\Rightarrow {\begin{array}{c}\frac{d^{2}y}{d{x}^2}\end{array}|}_{t=\frac{1}{2}}=-2\times \frac{4}{3}=-\frac{8}{3}$