Question

If $x=\sqrt{2^{{\text{cosec}}^{-1}t}}$ and $y=\sqrt{2^{{\sec }^{-1}t}}$ $(|t|\ge 1),$ then $\frac{dy}{dx}$ is equal to

• A. $\frac{y}{x}$
• B. $-\frac{y}{x}$
• C. $-\frac{x}{y}$
• D. $\frac{x}{y}$

Answer 1

The correct option is B $-\frac{y}{x}$

$x=\sqrt{2^{{\text{cosec}}^{-1}t}}\Rightarrow x^2=2^{{\text{cosec}}^{-1}t}\text{and,}y=\sqrt{2^{{\sec }^{-1}t}}\Rightarrow y^2=2^{{\sec }^{-1}t}\Rightarrow {\log }_2{x}^2+{\log }_2{y}^2={\text{cosec}}^{-1}t+{\sec }^{-1}t\Rightarrow {\log }_2{x}^2{y}^2=\frac{\pi }{2}\Rightarrow x^2{y}^2=2^{\pi /2}$
Differentiating w.r.t. $x$
$\Rightarrow 2x{y}^2+2x^{2}y{y}^{\prime }=0\Rightarrow \frac{dy}{dx}=-\frac{y}{x}$