Find dy/dx where x =√1-t^2 & y = sin^-1 t

Question

Find dy/dx where \(x = \sqrt{1-t^2} \,and\, y = sin^{-1} t\)

Answer 1

x = \(\sqrt{1-t^2},y=sin^{-1}(t) \)

\(\frac{dx}{dt}=\frac1{2\sqrt{1-t^2}}\times-2t\)  and \(\frac{dy}{dt} = \frac1{\sqrt{1-t^2}}\)

\(\frac{dy}{dx}=\cfrac{\frac{dy}{dt}}{\frac{dx}{dt}}\) = \(\cfrac{\frac{1}{1-t^2}}{\frac{-t}{\sqrt{1-t^2}}}\) = \(-\frac1t\)