#### explain solution RD Sharma class 12 chapter Differentiation exercise 10.4 question 16 maths

$(1+x)^{2} \frac{d y}{d x}+1=0$

Hint:

Use quotient rule and algebraic identities

Given:

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

Solution:

\begin{aligned} &x \sqrt{1+y}+y \sqrt{1+x}=0 \\ &x \sqrt{1+y}=-y \sqrt{1+x} \end{aligned}

Squaring both the side,

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

\begin{aligned} &x^{2}(1+y)=y^{2}(1+x) \\ &x^{2}+x^{2} y=y^{2}+x y^{2} \end{aligned}

$x^{2}-y^{2}=x y^{2}-x^{2} y$

$(x+y)(x-y)=x y(y-x)$                        $\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$

\begin{aligned} &(x+y)=-x y \\ &y+x y=-x \\ &y(1+x)=-x \end{aligned}

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

Differentiate this above equation w.r.t x

$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{-x}{1+x}\right)$

$\frac{d y}{d x}=\frac{(1+x) \cdot \frac{d(-x)}{d x}-(-x) \cdot \frac{d(1+x)}{d x}}{(1+x)^{2}}$                [Using quotient rule]

$\frac{d y}{d x}=\frac{-(1+x)+x(0+1)}{(1+x)^{2}}$                                $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

\begin{aligned} &\frac{d y}{d x}=\frac{-(1+x)+x}{(1+x)^{2}} \\ &\frac{d y}{d x}=\frac{-1-x+x}{(1+x)^{2}} \end{aligned}

$\frac{d y}{d x}=-\frac{1}{(1+x)^{2}}$

$(1+x)^{2} \frac{d y}{d x}=-1$

$(1+x)^{2} \frac{d y}{d x}+1=0$

Hence proved