Question

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

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

Answer 1

Answer: (A)

$-\frac{8}{3}$

The given equations are:

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

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

Differentiating once w.r.t to t we get,

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

$\Rightarrow \frac{dy}{dt}=\frac{-2t}{(1-t^2)}$

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

Differentiating again w.r.t to t according to the following equation we get,

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-2\left(\frac{\sqrt{1-t^2}+\frac{t^2}{\sqrt{1-t^2}}}{1-t^2}\right).\sqrt{1-t^2}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=2\left(\frac{1-t^2+t^2}{(1-t^2)}\right)$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-2}{1-t^2}$

The value of $\frac{d^{2}y}{d{x}^2}$ at t = 1/2 is

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-8}{3}$ ....Answer