Question

if x√1+y + y√1+x = 0
then find (1+x)² dy/dx​

Answer 1

Step-by-step explanation:

We have,

$x \sqrt{1 + y} + y \sqrt{1 + x} = 0$

$\implies \: x \sqrt{1 + y} = - y \sqrt{1 + x}$

$\implies \: \frac{ \sqrt{y + 1}}{ \sqrt{x + 1} } = - \frac{y}{x}$

$\implies \: \bigg( \frac{ \sqrt{y + 1}}{ \sqrt{x + 1} } \bigg)^{2} = \bigg(- \frac{y}{x} \bigg)^{2}$

$\implies \: \frac{ y + 1}{x + 1 } = \frac{y ^{2} }{x^{2} }$

$\implies \: ({x}^{2}) y + {x}^{2} = x ({y}^{2} )+ {y}^{2}$

Differentiating both sides w.r.t x,

$\implies \: 2x y + {x}^{2} \frac{dy}{dx} + 2x = {y}^{2} + 2xy \frac{dy}{dx} + 2y \frac{dy}{dx}$

$\implies \: ({x}^{2} - 2xy - 2y) \frac{dy}{dx} = {y}^{2} - 2xy - 2x$

$\implies \: \frac{dy}{dx} = \frac{{y}^{2} - 2xy - 2x }{ {x}^{2} - 2xy - 2y }$

Answer 2

Answer:

$\boxed{\sf \: \sf \: (1+x)^2\dfrac{dy}{dx} = \: - \: 1 \: }$

Step-by-step explanation:

Given that,

$\sf \: x \sqrt{1 + y} + y \sqrt{1 + x} = 0$

$\sf \: x \sqrt{1 + y} = - y \sqrt{1 + x}$

On squaring both sides, we get

$\sf \: (x \sqrt{1 + y})^{2} = (- y \sqrt{1 + x})^{2}$

$\sf \: {x}^{2}(1 + y) = {y}^{2}(1 + x)$

$\sf \: {x}^{2} + y {x}^{2} = {y}^{2} + x {y}^{2}$

$\sf \: {x}^{2} - {y}^{2} = x {y}^{2} - {x}^{2}y$

$\sf \: (x + y)(x - y)= xy(y - x)$

$\sf \: (x + y)(x - y)= - xy(x - y)$

$\sf \: x + y= - xy$

$\sf \: x= - xy - y$

$\sf \: x= - y(x + 1)$

$\sf \: y = - \: \dfrac{x}{x + 1}$

On differentiating both sides w. r. t. x, we get

$\sf \: \dfrac{d}{dx} y =\dfrac{d}{dx}\left( \dfrac{ - x}{x + 1}\right)$

$\sf \: \dfrac{dy}{dx} = \dfrac{(x + 1)\dfrac{d}{dx}( - x) - ( - x)\dfrac{d}{dx}(x + 1)}{ {(x + 1)}^{2} }$

$\sf \: \dfrac{dy}{dx} = \dfrac{(x + 1)( - 1) + x(1 + 0)}{ {(x + 1)}^{2} }$

$\sf \: \dfrac{dy}{dx} = \dfrac{ - x - 1 + x}{ {(x + 1)}^{2} }$

$\implies\sf \: \sf \: (1+x)^2\dfrac{dy}{dx} = - 1$

$\rule{190pt}{2pt}$

Formulae used:

$\sf \: \dfrac{d}{dx}\dfrac{u}{v} = \dfrac{u\dfrac{d}{dx}v - v\dfrac{d}{dx}u}{ {v}^{2} }$

$\sf \: \dfrac{d}{dx}x = 1$

$\sf \: \dfrac{d}{dx}k = 0$