Question

find dy/dx when x=a1-t²/1+t²,6=b2t/1+t²​

Answer 1

Step-by-step explanation:

dy/dt= d/dt( b2t/1+t²)

= b2[ (1+t²) - 2t²/]( 1+t²)²

= b2( 1-t²)/ ( 1+t²)²................(1)

dx/dt= a1[ (1+t²) ( -2t) - (1-t²) ( 2t)]/ (1+t²)²

= a1[ -2t-2t³-2t+2t³] / (1+t²)²

a1 ( -4t) / (1+t²)²..............(2)

(1) / (2)

(dy/dt)/ (dx/dt). =

dy/dx=b2[ (1-t²) / (-4t)]/a1

b2[ t/4 - 1/4t]/a1