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 $-\frac{y}{x}$

$\frac{dy}{dt}=e^{{\sin }^{-1}(t^2-1)}\times \frac{1}{\sqrt{1-(t^2-1{)}^2}}\times 2t$

$\frac{dx}{dt}=e^{{\sec }^{-1}\left(\frac{1}{t^2-1}\right)}\times \frac{1}{\left(\frac{1}{t^2-1}\right)\sqrt{{\left(\frac{1}{t^2-1}\right)}^2-1}}\times \frac{-2t}{(t^2-1{)}^2}$
$\frac{dx}{dt}=e^{{\sec }^{-1}\left(\frac{1}{t^2-1}\right)}\times \frac{1}{\left(\frac{1}{t^2-1}\right)\sqrt{\frac{1-(t^2-1{)}^2}{(t^2-1{)}^2}}}\times \frac{-2t}{(t^2-1{)}^2}$
$\frac{dx}{dt}=e^{{\sec }^{-1}\left(\frac{1}{t^2-1}\right)}\times \frac{(t^2-1{)}^2}{\sqrt{1-(t^2-1{)}^2}}\times \frac{-2t}{(t^2-1{)}^2}$

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


Alternate:
$y=e^{{\sin }^{-1}(t^2-1)}$
$x=e^{{\sec }^{-1}\left(\frac{1}{t^2-1}\right)}=e^{{\cos }^{-1}(t^2-1)}$
Now, $xy=e^{{\sin }^{-1}(t^2-1)}\cdot e^{{\cos }^{-1}(t^2-1)}$
$=e^{{\sin }^{-1}(t^2-1)+{\cos }^{-1}(t^2-1)}=e^{\pi /2}$
i.e, $xy=e^{\pi /2}$
Differentiating wrt $x,$ we get
$y+x\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{-y}{x}$