Use implicit derivative to determine \(dy/dx\).
\(\sqrt{xy} = {1+x^2y}\)
Note that √(xy) = x^(1/2) * y^(1/2)
So we have
x^(1/2) * y^(1/2) = 1 + x^2y
Using implicit differentiation
(1/2)x^(-1/2)* y^(1/2) + x^(1/2) * (1/2) y^(-1/2) y' = 0 + 2xy + x^2 y' =
y^(1/2) / [ 2x^(1/2) ] + (x^(1/2) / [2 y ^(1/2) ]) y' = 2xy + x^2 y'
Gather the y' terms on the left and everything else on the right we have that
y' [ x^(1/2)/ [ 2y^(1/2)] - x^2 ] = 2xy - y^(1/2) / [ 2x^(1/2) ]
y' = dy/dx = ( 2xy - y^(1/2) / [ 2x^(1/2) ] ) / [ x^(1/2)/ [ 2y^(1/2)] - x^2 ]