If ( y=e^ ⁻¹ (t²-1 ) & x=e^ ⁻¹ ( 1 t²-1 ) ) then ( d y d x ) is equal to

Step-by-step solution

Given that, y=e^ ⁻¹ (t²-1 ) and x=e^ ⁻¹ ( 1 t²-1 ) So, y= ⁻¹ (t²-1 ) and x= ⁻¹ ( 1 t²-1 )= ⁻¹ (t²-1 ) Now, y+ x= ⁻¹ (t²-1 )+ ⁻¹ (t²-1 )= 2 y+ x= 2 Now, differentiate with respect to x , we get 1 y d y d x + 1 x =0 d y d x = -y x