Question

if

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

options are :-
$a) \dfrac{x+1}{x}$

$b) \dfrac{ 1}{1+x}$
$c) \dfrac{-1}{(1+x)²}\\ \\ d) \dfrac{x}{1+x}$
​

Answer 1

We have given that :

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

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

$\bigstar \: \pmb{ \mathfrak{ \text Squaring \: both \: sides}}$

$: \longmapsto \rm {x}^{2} (1 + y) = ( - y) ^{2} (1 + x)$

$: \longmapsto \rm {x}^{2} (1 + y) = ({ - y)}^{2} (1 + x)$

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

$: \longmapsto \rm {x}^{2} + {x}^{2} y = {y}^{2} + xy {}^{2}$

$: \longmapsto \rm {x}^{2} - {y}^{2} = {xy}^{2} - {x}^{2} y$

$: \longmapsto \rm (x + y)(x - y) = xy(y - x)$

$: \longmapsto \rm - (x + y) \: \cancel{(y - x)} = xy \: \cancel{(y - x)}$

$: \longmapsto \rm - x - y = xy$

$: \longmapsto \rm - x = xy + y$

$: \longmapsto \rm - x = y(x + 1)$

$: \longmapsto \rm y = - \frac{x}{x + 1}$

$\bigstar \: \pmb{\frak{Differentiating \: both \: sides \: \text w.r.t. \: \: \text x}}$

$: \longmapsto \rm \frac{dy}{dx} = \frac{d}{dx} \bigg( - \frac{x}{x + 1} \bigg)$

$: \longmapsto \rm \frac{dy}{dx} = - \frac{d}{dx} \bigg( \frac{x}{x + 1} \bigg)$

$\bigstar \: \pmb{ \frak{Differentiating \: using \: quotient \: rule}}$

$: \longmapsto \rm \frac{dy}{dx} = - \bigg( \frac{(x + 1) \frac{d}{dx} ( x) - x \frac{d}{dx} (x + 1)}{ {(x + 1)}^{2} } \bigg)$

$: \longmapsto \rm \frac{dy}{dx} = - \bigg( \frac{(x + 1).1 - x.1}{ {(1 + x)}^{2} } \bigg)$

$: \longmapsto \rm \frac{dy}{dx} = - \bigg( \frac{x + 1 - x}{ {(1 + x)}^{2} } \bigg)$

$\large: \longmapsto \underbrace{\underline{\bf \frac{dy}{dx} = - \frac{1}{ {(1 + x)}^{2} }}}$

Hence, option C is correct.