If x= ⁻¹ 1+ t ^2 -1 t , y = ⁻¹ 1- t ^2 1+ t ^2 , then dy d x is equal to

Step-by-step solution

Find dy dx using parametric differentiation with substitution. Let t = , so = ⁻¹t . For x = ⁻¹ 1+t^2 -1 t , substitute to obtain x = ⁻¹ -1 which simplifies to x = ⁻¹ ( /2) = 2 = 1 2 ⁻¹t . Differentiate: dx dt = 1 2(1+t^2) . For y = ⁻¹ 1-t^2 1+t^2 , substitution gives y = ⁻¹( (2 )) where = ⁻¹t so y = 2 ⁻¹t and dy dt = 2 1+t^2 . Apply the chain rule: dy dx = dy/dt dx/dt = 2/(1+t^2) 1/(2(1+t^2)) = 4 . The derivative is constant, dy dx = 4 .