Question

If $y=e^{{\sin }^{-1}(t^2-1)}$ and $x=e^{{\sec }^{-1}\left(\frac{1}{t^2-1}\right)}$ then $\frac{dy}{dx}$ is equal to

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

Answer 1

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

We have, $y=e^{{\sin }^{-1}(t^2-1)}$ and $x=e^{{\sec }^{-1}\left(\frac{1}{t^2-1}\right)}=e^{{\cos }^{-1}(t^2-1)}$, $[\because {\cos }^{-1}x={\sec }^1\frac{-1}{x}]$

Now multiply $x$ and $y$,

$xy=e^{{\sin }^{-1}(t^2-1)}\cdot e^{{\cos }^{-1}(t^2-1)}=e^{{\sin }^{-1}(t^2-1)+{\cos }^{-1}(t^2-1)}=e^{\pi /2}$ $[\because {\sin }^{-1}\theta +{\cos }^{-1}\theta =\frac{\pi }{2}]$

$\Rightarrow xy=\text{constant}$

Differentiate both sides w.r.t. $x$

$\Rightarrow y+x\frac{y}{dx}=0$

$\therefore \frac{dy}{dx}=-\frac{y}{x}$