If `xsqrt(1+y)+ysqrt(1+x)=0`, for, `-1

Options

A. ` (1)/((1 + x))`
B. `(1)/((1 + x)^(2))`
C. ` (1)/(1 + x^(2))`
D. `-(1)/((1 + x)^(2))`

Answer

D

Explanation

Given , ` x sqrt( 1 + x) + y sqrt(1 + x) = 0 `
`rArr x sqrt(1 + y) = - y sqrt(1 + x)`
On squaring both sides of Eq. (i) , we get
` x^(2) (1 + y) = y^(2) (1 +x)`
`rArr x^(2) - y^(2) + x^(2) y - y^(2) x = 0 `
` rArr (x-y) (x+y) + xy (x -y) = 0 `
`rArr (x-y) {x +y + xy} = 0 `
`rArr x - y = 0 or x + y + xy = 0 `
` rArr y = x or y. (1 + x) = - x `
` rArr y = x or y = (-x)/(1 + x)`
but y = x does not satisfy the given equation .
So, we consider only ` y = - (x)/(1 + x)`
On differentiating both sides w.r.t.x, we get
`(dy)/(dx) = (d)/(dx) ((-x)/(1 + x)) = - ((1 +x) (d)/(dx) (x) - x(d)/(dx) (1 + x))/((1 +x)^(2))`
` = - ((1 + x) .1 - x(0+1))/((1 + x)^(2)) = - (1)/((1 + x)^(2))`