If \( x\sqrt{1+y}+y\sqrt{1+x}=0,x\ne y \), then \( \displaystyle\frac{dy}{dx} \) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( -\displaystyle\frac{1}{(1+x)^{2}} \)
Here y can be found explicitly. Moving one term across and squaring:
\( x\sqrt{1+y}=-y\sqrt{1+x} \)
\( x^{2}(1+y)=y^{2}(1+x) \)
\( x^{2}-y^{2}+x^{2}y-xy^{2}=0 \)
Factorising:
\( (x-y)(x+y)+xy(x-y)=0 \)
\( (x-y)(x+y+xy)=0 \)
Since \( x\ne y \), we get \( x+y+xy=0 \), so
\( y(1+x)=-x \)
\( \displaystyle y=-\frac{x}{1+x} \)
By the quotient rule:
\( \displaystyle \frac{dy}{dx}=-\frac{(1+x)\cdot 1-x\cdot 1}{(1+x)^{2}} \)
\( \displaystyle =-\frac{1}{(1+x)^{2}} \)
NCERT (2026-27 edition):
Continuity and Differentiability
