Practicing Success
Practicing Success
CUET Preparation Today
If $x\sqrt{1+y}+y\sqrt{1+x}=0, $ then $\frac{dy}{dx}=$ |
$\frac{1}{(1+x)^2}$ $\frac{-1}{(1+x)^2}$ $(1+x)^2$ $-(1+x)^2$ |
$\frac{-1}{(1+x)^2}$ |
