Question

x=sqrt(1-t^2) ; y=sin^-t
$x = \sqrt{1 - t {}^{2} } \: \: \: y = sin { }^{ - t}$
find dy/dx of the above.​

Answer 1

Answer:

qrt(1-t^2) ; y=sin^-t

$x = \sqrt{1 - t {}^{2} } \: \: \: y = sin { }^{ - t}$

find dy/dx of the above