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If $x \sqrt {1+y}+y \sqrt {1+x}=0$ then $\large\frac{dy}{dx}$ is equal to :

$\begin {array} {1 1} (a)\;\frac{1}{(1+x)^2} & \quad (b)\;-\frac{1}{(1+x)^2} \\ (c)\;\frac{1}{1+x^2} & \quad (d)\;\frac{1}{1-x^2} \end {array}$

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

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