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Question Number 20831 by j.masanja06@gmail.com last updated on 04/Sep/17

i f x \sqrt{1 + y} + y \sqrt{1 + x} = 0 p r o v e t h a t

\frac{d y}{d x} = - {(1 + x)}^{- 2}

Answered by ajfour last updated on 04/Sep/17

x \sqrt{1 + y} = - y \sqrt{1 + x}

s o

x^{2} (1 + y) = y^{2} (1 + x)

o r x^{2} - y^{2} = x y (y - x)

\Rightarrow i f y \neq x, x + y + x y = 0

o r y = - \frac{x}{1 + x}

d i f f e r e n t i a t i n g w h i c h w e g e t

\frac{d y}{d x} = - \frac{[(1 + x) - x]}{{(1 + x)}^{2}} = - {(1 + x)}^{- 2} .

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