If $\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0, x
eq y$, then $\frac{d y}{d x}=0$

Question

If $\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0, x 
eq y$, then $\frac{d y}{d x}=0$

Aicharya · Accepted Answer

$\begin{array}{l} \frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0 \ x \cdot(\sqrt{1+y})+y(\sqrt{1+x})=0 \quad \Rightarrow x \cdot \sqrt{1+y}=-y \sqrt{1+x} \end{array}$ Squaring both sides we get, $\begin{array}{l} x^{2}(1+y)=y^{2}(1+x) \Rightarrow x^{2}+x^{2} y=y^{2}+x y^{2} \ x^{2}-y^{2}=x y^{2}-x^{2} y \Rightarrow(x-y)(x+y)=-x y(x-y) \ x+y=-x y \Rightarrow(1+x) y=-x \Rightarrow y=\frac{-x}{1+x} \end{array}$ Differentiating both sides w.r.t. $x$, we get $\begin{aligned} & \frac{d y}{d x}=-\frac{(1+x)(1)-(x)(1)}{(1+x)^{2}}=\frac{-1}{(1+x)^{2}} \ 	herefore &(1+x)^{2} \frac{d y}{d x}=-1 \end{aligned}$

If $\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0, x \neq y$, then $\frac{d y}{d x}=0$

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