question_answer

If \[x\sqrt{1+y}+y\,\,\sqrt{1+x}=0\] and \[x\ne y,\] prove that \[\frac{dy}{dx}=\frac{1}{{{(x+1)}^{2}}}.\]

Answer:

Given, \[x\sqrt{1+y}+y\sqrt{1+x}=0\] \[\Rightarrow \]   \[x\sqrt{1+y}=-\,y\sqrt{1+x}\] \[\Rightarrow \]   \[{{x}^{2}}(1+y)={{y}^{2}}(1+x)\][squaring both sides] \[\Rightarrow \]   \[{{x}^{2}}-{{y}^{2}}={{y}^{2}}x-{{x}^{2}}y\] \[\Rightarrow \] \[(x+y)(x-y)=-xy(x-y)\] \[\Rightarrow \] \[x+y=-\,xy\] \[\Rightarrow \]   \[x=-y-xy\] \[\Rightarrow \] \[y(1+x)=-x\] \[\Rightarrow \]   \[y=-\frac{x}{1+x}\] \[\therefore \]      \[\frac{dy}{dx}=-\left[ \frac{(1+x)\cdot 1-x(0+1)}{{{(1+x)}^{2}}} \right]=-\frac{x}{{{(1+x)}^{2}}}\]          Hence proved.