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If $x \sqrt{1+y}+y \sqrt{1+x}=0$, then $\frac{d y}{d x}$ is equal to
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Verified Answer
The correct answer is:
$-\frac{1}{(1+x)^2}$
Given that
or
$x \sqrt{1+y}=-y \sqrt{1+x}$
On squaring both sides, we get
$\begin{array}{cc}
& x^2(1+y)=y^2(1+x) \\
\Rightarrow & x^2-y^2+x^2 y-x y^2=0 \\
\Rightarrow & (x-y)(x+y)+x y(x-y)=0 \\
\Rightarrow & (x-y)(x+y+x y)=0
\end{array}$
$x-y \neq 0$ because it does not satisfy the Eq. (i)
$\begin{aligned}
& \therefore \quad x+y+x y=0 \\
& \Rightarrow \quad y=-\frac{x}{1+x} \\
&
\end{aligned}$
On differentiating w.r.t. $x$, we get
$\begin{aligned}
\frac{d y}{d x} & =-\frac{(1+x)(1)-x(1)}{(1+x)^2} \\
& =-\frac{1}{(1+x)^2}
\end{aligned}$
or
$x \sqrt{1+y}=-y \sqrt{1+x}$
On squaring both sides, we get
$\begin{array}{cc}
& x^2(1+y)=y^2(1+x) \\
\Rightarrow & x^2-y^2+x^2 y-x y^2=0 \\
\Rightarrow & (x-y)(x+y)+x y(x-y)=0 \\
\Rightarrow & (x-y)(x+y+x y)=0
\end{array}$
$x-y \neq 0$ because it does not satisfy the Eq. (i)
$\begin{aligned}
& \therefore \quad x+y+x y=0 \\
& \Rightarrow \quad y=-\frac{x}{1+x} \\
&
\end{aligned}$
On differentiating w.r.t. $x$, we get
$\begin{aligned}
\frac{d y}{d x} & =-\frac{(1+x)(1)-x(1)}{(1+x)^2} \\
& =-\frac{1}{(1+x)^2}
\end{aligned}$
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