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