Question

If x√1+y + y√1+X = 0 prove that dy/dx = -1/(1+x)^2

Answer 1

Hey there!

Solution:


We have $x\sqrt{1 + y} + y\sqrt{1 + x} = 0\\ \\ x\sqrt{1 + y} = -y\sqrt{1 + x} \\ \\ Now \:Squaring\: Both\: Sides\\ \\ x^{2} ( 1 + y ) = y^{2}( 1 + x ) \\ \\ x^{2} + yx^{2} = y^{2} + y^{2}x \\ \\ x^{2} + yx^{2} - y^{2} - yx^{2} = 0\\ \\ ( x^{2} - y^{2} ) + xy( x - y ) = 0\\ \\ ( x - y ) ( x + y ) + ( xy )( x - y ) = 0\\ \\ Taking\: ( x - y )\: common \:from\: the \:L.H.S\\ \\ ( x - y ) ( x + y + xy ) = 0\\ \\ We \:know\: that\: x - y = 0\: is \:not \:possible \:because \:it\: implies\: x = y \\ \\ x + y + xy = 0 \\ \\ y + xy = -x \\ \\ y ( 1 + x ) = 0 \\ \\ y = -\dfrac{x}{1+x} \\ \\ Now \:differentiate\: both\: sides\\ \\ \dfrac{dy}{dx} = \dfrac{(1 + x ).\dfrac{d(-x)}{dx} - (-x).\dfrac{d(1 + x)}{dx}}{( 1 + x )^{2}} \\ \\ = \dfrac{(1 + x)(-1) + x(1)}{(1 + x )^{2}}\\ \\ = \dfrac{-1}{( 1 + x )^{2}}$


Hence Proved.


#Be Brainly.

Answer 2

Your Answer!!

$\begin{gathered}x\sqrt{1 + y} + y\sqrt{1 + x} = 0\\ \\ x\sqrt{1 + y} = -y\sqrt{1 + x} \\ \\ \:Squaring\: Both\: Sides\\ \\ x^{2} ( 1 + y ) = y^{2}( 1 + x ) \\ \\ x^{2} + yx^{2} = y^{2} + y^{2}x \\ \\ x^{2} + yx^{2} - y^{2} - yx^{2} = 0\\ \\ ( x^{2} - y^{2} ) + xy( x - y ) = 0\\ \\ ( x - y ) ( x + y ) + ( xy )( x - y ) = 0\\ \\ ( x - y ) ( x + y + xy ) = 0 \\ \\ x + y + xy = 0 \\ \\ y + xy = -x \\ \\ y ( 1 + x ) = 0 \\ \\ y = -\dfrac{x}{1+x} \\ \\ \dfrac{dy}{dx} = \dfrac{(1 + x ).\dfrac{d(-x)}{dx} - (-x).\dfrac{d(1 + x)}{dx}}{( 1 + x )^{2}} \\ \\ = \dfrac{(1 + x)(-1) + x(1)}{(1 + x )^{2}}\\ \\ = \dfrac{-1}{( 1 + x )^{2}}\end{gathered}$

${\boxed{Hence\:Proved}}$