Question

If $y=e^{\sin ^{-1}\left(t^2-1\right)}\,\&\, x=e^{\sec ^{-1}\left(\frac{1}{t^2-1}\right)}$ then $\frac{d y}{d x}$ 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, $ x y=e^{\sin ^{-1}\left(t^{2}-1\right)} \cdot e^{\sec ^{-1}\left(\frac{1}{t^{2}-1}\right)} $
$\Rightarrow x y=e^{\sin ^{-1}\left(t^{2}-1\right)+\cos ^{-1}\left(t^{2}-1\right)}$
${\left[\because \sec ^{-1}\left(\frac{1}{t^{2}-1}\right)=\cos ^{-1}\left(t^{2}-1\right)\right]} $
$\Rightarrow x y=e^{\pi / 2} \,\,\left[\because \sin ^{-1} \theta+\cos ^{-1} \theta=\frac{\pi}{2}\right]$
On differentiating both sides w.r.t. $x$, we get
$x \frac{d y}{d x}+y=0$
$\Rightarrow \frac{d y}{d x}=\frac{y}{x}$