x1 + y + y1 + x = 0, then dy dx = - Vidyadip

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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