Question

If `xsqrt(1 + y) + ysqrt(1 + x)` = 0 and x ≠ y, then prove that `"dy"/"dx" = - 1/(x + 1)^2`

Answer 1

Given `xsqrt(1 + y) + ysqrt(1 + x)` = 0

`xsqrt(1 + y) = - ysqrt(1 + x)`

Squaring both sides we get

⇒ x² (1 + y) = y² (1 + x)

⇒ x² + x²y = y² + y²x

⇒ x² – y² + x²y – y²x = 0

⇒ (x + y) (x – y) + xy(x – y) = 0

⇒ (x – y) [(x + y) + xy] = 0

∴ x – y = 0 (or) x + y + xy = 0

x = y (or) x + y + xy = 0

Given that x ≠ y

x + y + xy = 0

⇒ y + xy = -x

⇒ y(1 + x) = -x

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

`"dy"/"dx" = - [((1 + x)1 - x(1 + 0))/(1 + x)^2]`

`= - [(1 + x - x)/(1 + x)^2]`

`= - [1/(1 + x)^2]`

`= - 1/(1 + x)^2`

Hence proved.