Q49 / 100 · BITSAT 2024 (Memory Based Paper 3) · Mathematics
If $x \sqrt{1+y}+y \sqrt{1+x}=0$, then $\frac{d y}{d x}=$
- A$\frac{x+1}{x}$
- B$\frac{1}{1+x}$
- Correct answerC$\frac{-1}{(1+x)^2}$
- D$\frac{x}{1+x}$
Solution
$\begin{aligned}
& \text { Given, } x \sqrt{1+y}+y \sqrt{1+x}=0 \\
& \Rightarrow \quad x \sqrt{1+y}=-y \sqrt{1+x} \\
& \Rightarrow \quad(x \sqrt{1+y})^2=(-y \sqrt{1+x})^2
\end{aligned}$
$\begin{array}{lr}
\Rightarrow & x^2(1+y)=y^2(1+x) \\
\Rightarrow & x^2-y^2=y^2 x-x^2 y \\
\Rightarrow & (x-y)(x+y)=x y(y-x) \\
\Rightarrow & x+y=-x y \\
\Rightarrow & x+y+x y=0 \\
\Rightarrow & y=\frac{-x}{1+x}=-1+\frac{1}{1+x}
\end{array}$
Differentiating w.r.t, $x$
$\frac{d y}{d x}=-\frac{1}{(1+x)^2}$
Hence, the correct answer is option: 'C'.