Question

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

Answer 1

`DIFFERENTIAL \: \: \: CALCULUS \\ \\ \\ Given \: expression \: - \\ \: \\ x √(1 + y) \: = \: y √(1 + x) \\ \\ x √(1 + y) \: = \: - y √(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 \: \: = \: \: ( - x)/(1 + x) \\ \\ (dy)/(dx) \: \: = \: - \: \frac{(1 + x) - x.1}{ {(1 + x)}^(2) } \\ \\ \\ (dy)/(dx) = \frac{ - 1}{( {1 + x)}^(2) } \: \: \: \: \: \: \: \: \: Ans.`