MCQ
$x\sqrt {1 + y} + y\sqrt {1 + x} = 0$, then ${{dy} \over {dx}} = $
- A$1 + x$
- B${(1 + x)^{ - 2}}$
- C$ - {(1 + x)^{ - 1}}$
- ✓$ - {(1 + x)^{ - 2}}$
$\Rightarrow$ $(x-y)(x+y+xy)=0$ $\Rightarrow$ $x+y+xy=0,$ $\,\,\,\left\{ \because x\ne y \right\}$
$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{{(1 + x)}^2}}}$.
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