Question

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

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

Answer 1

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

$x=\sqrt{2^{cose{c}^{-1}t}}\text{and}y=\sqrt{2^{{\sec }^{-1}t}}\Rightarrow {\log }_2{x}^2+{\log }_2{y}^2=\frac{\pi }{2}(\because {\text{cosec}}^{-1}t+{\sec }^{-1}t=\frac{\pi }{2})\Rightarrow x^2{y}^2=c\Rightarrow 2x{y}^2+2x^{2}y{y}^{\prime }=0\Rightarrow \frac{dy}{dx}=-\frac{y}{x}$