Question

If x√1+ y. + y√ 1+x = 0
then ( 1+ x² ) dy/dx= ?​

Answer 1

Seems like yourquestion is 'find (1 + x)²dy/dx'.

Given, x√(1 + y) = - y√(1 + x)

Square both sides,

=> x²(1 + y) = y²(1 + x)

=> x² + x²y = y² + y²x

=> x² - y² = y²x - x²y

=> (x + y)(x - y) = - xy(x - y)

=> x + y = - xy

=> y + xy = - x

=> y(1 + x) = - x

=> y = - x/(1 + x)

Differentiate both sides wrt x:

=> y' = - [1(1 + x) - x(0 + 1)] / (1 + x)²

=> y' = - 1/(1 + x)²

=> (1 + x)² y' = - 1

Hence, (1 + x)²y' = - 1