Question

If $y=e^{{\sin }^{-1}\left(t^2-1\right)}\&x=e^{{\sec }^{-1}\left(\frac{1}{t^2-1}\right)}$ 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

Answer: (B)

We have, $y=e^{{\sin }^{-1}\left(t^2-1\right)}$ and
$x=e^{{\sec }^{-1}\left(\frac{1}{t^2-1}\right)}$
Now, $xy=e^{{\sin }^{-1}\left(t^2-1\right)}\cdot e^{{\sec }^{-1}\left(\frac{1}{t^2-1}\right)}$
$\Rightarrow xy=e^{{\sin }^{-1}\left(t^2-1\right)+{\cos }^{-1}\left(t^2-1\right)}$
$\left[\because {\sec }^{-1}\left(\frac{1}{t^2-1}\right)={\cos }^{-1}\left(t^2-1\right)\right]$
$\Rightarrow xy=e^{\pi /2}\left[\because {\sin }^{-1}\theta +{\cos }^{-1}\theta =\frac{\pi }{2}\right]$
On differentiating both sides w.r.t. $x$, we get
$x\frac{dy}{dx}+y=0$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}$