If ( x= 2 t 1+t² , y= 2 t 1-t² ), then ( d y d x ) is equal to

Step-by-step solution

Given that, ( x= 2 t 1+t² ) and ( y= 2 t 1-t² ) So, ( x= ⁻¹ ( 2 t 1+t² ) (1) ) and ( y= ⁻¹ ( 2 t 1-t² ) (2) ) We know that ( ⁻¹ ( 2 t 1+t² )=2 ⁻¹ t ) and ( ⁻¹ ( 2 t 1-t² )=2 ⁻¹ t ) So, Eqs. (1) and (2) becomes ( x=2 ⁻¹ t ) and ( y=2 ⁻¹ t ) Differentiating Eqs. (1) and (2) with respect to t, we get ( d y d x = d ( 2 ⁻¹ d ) d (2 ⁻¹ ) d =1 )