If x√(1+y) + y√(1+x) = 0 , then dy/dx = ?

[tex]DIFFERENTIAL \: \: \: CALCULUS \\ \\ \\ Given \: expression \: - \\ \: \\ x \sqrt{1 + y} \: = \: y \sqrt{1 + x} \\ \\ x \sqrt{1 + y} \: = \: - y \sqrt{1 + x} \\ \\ \\ Squaring \: both \: sides \: , \: We \: get \: - \: \\ \\ {x}^{2} (1 + y) = {y}^{2} (1 + x) \\ \\ {x}^{2} + {x}^{2} y - {y}^{2} - x {y}^{2} = 0 \\ \\ ( {x}^{2} - {y}^{2} ) + xy(x - y) = 0 \\ \\ (x - y)(x + y + xy) = 0 \\ \\ Therefore \: , \: either \: \: y=x \: \: or \: \\ \: \: \: \: \: \: \: \: \: \: \: x+y+xy=0 \\ \\ \\ Since \: , \: \: x=y \: \: doesn't \: satisfy \: the \\ given \: function \: , \: we \: reject \: it \\ \\ x + y + xy = 0 \\ \\ y \: \: = \: \: \frac{ - x}{1 + x} \\ \\ \frac{dy}{dx} \: \: = \: - \: \frac{(1 + x) - x.1}{ {(1 + x)}^{2} } \\ \\ \\ \frac{dy}{dx} = \frac{ - 1}{( {1 + x)}^{2} } \: \: \: \: \: \: \: \: \: Ans.[/tex]