If $x\sqrt {1 + y} + y\sqrt {1 + x} = 0,$ $\frac{{dy}}{{dx}}$ equals

$\frac{{ - 1}}{{(1 + x)}} + \frac{1}{{{{(1 + x)}^2}}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

x\sqrt {1 + y} + y\sqrt {1 + x} = 0,

\frac{{dy}}{{dx}}

equals

View answer and explanation

Correct option: B

\frac{{ - 1}}{{{{(1 + x)}^2}}}

Option B (

\frac{{ - 1}}{{{{(1 + x)}^2}}}

) is correct because it is the result of differentiating the simplified equation after eliminating y.

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v = \left( {\frac{x}{y}} \right),

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}