If x 1+y +y 1+x =0, for, < x < 1, prove that - Vidyadip

Question

$\text{If x}\sqrt{1+\text{y}}+\text{y}\sqrt{1+\text{x}}=0,$ for, < x < 1, prove that

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Answer

It is given that,
$\text{x}\sqrt{1+\text{y}}+\text{y}\sqrt{1+\text{x}}=0$
$\Rightarrow\ \text{x}\sqrt{1+\text{y}}=-\text{y}\sqrt{1+\text{x}}$
Squaring both sides, we obtain
$\text{x}^2(1+\text{y})=\text{y}^2(1+\text{x})$
$\Rightarrow\ \text{x}^2+\text{x}^2\text{y}=\text{y}^2+\text{xy}^2$
$\Rightarrow\ \text{x}^2-\text{y}^2=\text{xy}^2-\text{x}^2\text{y}$
$\Rightarrow\ \text{x}^2-\text{y}^2=\text{xy}(\text{y}-\text{x})$
$\Rightarrow\ (\text{x}+\text{y})(\text{x}-\text{y})=\text{xy}(\text{y}-\text{x})$
$\therefore\ \text{x}+\text{y}=-\text{xy}$
$\Rightarrow\ (1+\text{x})\text{y}=-\text{x}$
$\Rightarrow\ \text{y}=\frac{-\text{x}}{(1+\text{x})}$
Differentiating both sides with respect to x, we obtain
$ \text{y}=\frac{-\text{x}}{(1+\text{x})}$
$\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{x})\frac{\text{d}}{\text{dx}}(\text{x)}-\text{x}\frac{\text{d}}{\text{dx}}(1+\text{x})}{(1+\text{x})^2}$ $=-\frac{(1+\text{x})-\text{x}}{(1+\text{x})^2}=-\frac{1}{(1+\text{x})^2}$
Hence, proved.

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