If $x \sqrt{1+y}+y \sqrt{1+x}=0$, then $\frac{d y}{d x}$ is equal to
- $\frac{1}{(1+x)^2}$
- $-\frac{1}{(1+x)^2}$
- $\frac{1}{1+x^2}$
- $\frac{1}{1-x^2}$
Solution
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 $\begin{aligned} \frac{d y}{d x} & =-\frac{(1+x)(1)-x(1)}{(1+x)^2} \\ & =-\frac{1}{(1+x)^2} \end{aligned}$
Asked in: AP EAMCET 2005