Question

If $x\sqrt{1+y}+y\sqrt{1+x}=0$, then show that $\frac{dy}{dx}=-\frac{1}{(1+x{)}^2}$.

Answer 1

We have,

$\begin{array}{c}x\sqrt{1+y}+y\sqrt{1+x}=0\\ \Rightarrow x\sqrt{1+y}=-y\sqrt{1+x}\\ \Rightarrow x^2\left(1+y\right)=y^2\left(1+x\right)\\ \Rightarrow x^2+x^{2}y=y^2=x{y}^2\\ \Rightarrow x^2-y^2=-xy\left(x-y\right)\\ \Rightarrow \left(x+y\right)\left(x-y\right)=-xy\left(x-y\right)\\ \Rightarrow \left(x+y\right)=-xy\\ \Rightarrow x=-y\left(x+1\right)\\ \Rightarrow -\frac{x}{\left(1+x\right)}=y\\ Then,differentiatewithrespecttoxbe,\\ \Rightarrow \frac{dy}{dx}=-\frac{\left(1+x\right)-x}{{\left(1+x\right)}^2}=-\frac{1}{{\left(x+1\right)}^2}ProvedAns.\end{array}$