MCQ
$x\sqrt {1 + y} + y\sqrt {1 + x} = 0$ માટે $\frac{{dy}}{{dx}} = ............$
- A$\frac{1}{{{{\left( {1 + x} \right)}^2}}}$
- ✓$\frac{{ - 1}}{{{{\left( {1 + x} \right)}^2}}}$
- C$\frac{1}{{1 + {x^2}}}$
- D$\frac{{ - 1}}{{1 + {x^2}}}$
$x\sqrt{1+y}+y\sqrt{1+x}=0$
$\therefore x^2(1+y)=Y^2(1+x)$
$\therefore x^2+x^2y=y^2+y^2x$
$\therefore x^2-y^2=xy(y-x)$
$\therefore (x-y)(x+y)=-xy(x-y)$
$\therefore x+y=-xy \Rightarrow x+y+xy=0$
$\therefore x+y(1+x)=0 \Rightarrow y=\frac{-x}{1+x}$
$\therefore \frac{dy}{dx}=\frac{(1+x)(-1)+x}{(1+x^2)}=\frac{-1-x+x}{(1+x)^2}$
$=\frac{-1}{(1+x)^2}$
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