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}}$

Answer

Correct option: D.
$ - {(1 + x)^{ - 2}}$
d
(d) $x\sqrt {1 + y} + y\sqrt {1 + x} = 0$ $\Rightarrow$  ${x^2}(1 + y) = {y^2}(1 + x)$

$\Rightarrow$ $(x-y)(x+y+xy)=0$ $\Rightarrow$ $x+y+xy=0,$    $\,\,\,\left\{ \because x\ne y \right\}$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{{(1 + x)}^2}}}$.

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