Question

If \( x\sqrt{1+y} +y\sqrt{1+x}=0, \: find \:\large \frac{dy}{dx} \)

Answer 1

Question:

$\sf \: If \: x \sqrt{1 + y} + y \sqrt{1 + x} = 0, \: find \: \dfrac{dy}{dx}$

Answer:

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

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

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

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

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

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

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

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

Formulae used:

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

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

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