If x 1+y +y 1+x =0, prove that (1+x)^2 dx dx +1=0 - Vidyadip

Question

If $\text{x}\sqrt{1+\text{y}}+\text{y}\sqrt{1+\text{x}}=0,$ prove that $(1+\text{x})^2\frac{\text{dx}}{\text{dx}}+1=0$

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Answer

We have $\text{x}\sqrt{1+\text{y}}+\text{y}\sqrt{1+\text{x}}=0$
$\Rightarrow\text{x}\sqrt{1+\text{y}}=-\text{y}\sqrt{1+\text{x}}$
Squaring both sides, we get,
$\Rightarrow\big(\text{x}\sqrt{1+\text{y}}\big)^2=(-\text{y}\sqrt{1+\text{x}}\big)^2$
$\Rightarrow\text{x}^2\big(1+\text{y}\big)=\text{y}^2(1+\text{x}\big)$
$\Rightarrow\text{x}^2+\text{x}^2\text{y}=\text{y}^2+\text{y}^2\text{x}$
$\Rightarrow\text{x}^2-\text{y}^2=\text{y}^2\text{x}-\text{x}^2\text{y}$
$\Rightarrow(\text{x}-\text{y})(\text{x}+\text{y})=\text{x}\text{y}(\text{y}-\text{x})$
$\Rightarrow(\text{x}+\text{y})=-\text{x}\text{y}$
$\Rightarrow\text{y}+\text{x}\text{y}=-\text{x}$
$\Rightarrow\text{y}(1+\text{x})=-\text{x}$
$\Rightarrow\text{y}=\frac{-\text{x}}{(1+\text{x})}$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\bigg[\frac{-(1+\text{x})\frac{\text{d}}{\text{dx}}(\text{x})-(-\text{x})\frac{\text{d}}{\text{dx}}(\text{x}+1)}{(1+\text{x})^2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big[\frac{-(1+\text{x})(1)+\text{x}(1)}{(1+\text{x})^2}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big[\frac{-1-\text{x}+\text{x}}{(1+\text{x})^2}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1}{(1+\text{x})^2}$
$\Rightarrow(1+\text{x})^2\frac{\text{dy}}{\text{dx}}=-1$
$\Rightarrow(1+\text{x})^2\frac{\text{dy}}{\text{dx}}+1=0$

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