Question

$x\sqrt{y+1} + y\sqrt{x+1} =0$
then find the value of dy/dx=----

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

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

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

On squaring both sides, we get

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

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

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

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

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

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

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

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

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

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

$\rm \: \dfrac{d}{dx} y \: = \: - \: \dfrac{d}{dx}\bigg(\dfrac{x}{x + 1}\bigg)$

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

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

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

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

Hence,

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

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

Formulae Used :-

$\boxed{ \rm{ \:\dfrac{d}{dx} \frac{u}{v} \: = \: \frac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } \: \: }}$

$\boxed{ \rm{ \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}$

$\boxed{ \rm{ \:\dfrac{d}{dx}k= 0 \: }}$

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

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$