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

Question

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

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Answer

Here,
$\log\sqrt{\text{x}^2+\text{y}^2}=\tan^{-1}\Big(\frac{\text{x}}{\text{y}}\Big)$
$\Rightarrow\log(\text{x}^2+\text{y}^2)^{\frac{1}{2}}=\tan^{-1}\Big(\frac{\text{x}}{\text{y}}\Big)$
$\Rightarrow\frac{1}{2}\log(\text{x}^2+\text{y}^2)=\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)$
Differentiating with respect to x,
$\Rightarrow\frac{1}{2}\frac{\text{d}}{\text{dx}}\log(\text{x}^2+\text{y}^2)=\frac{\text{d}}{\text{dx}}\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow\frac{1}{2}\times\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y}^2)=\frac{1}{1+\Big(\frac{\text{y}}{\text{x}}\Big)^2}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow\frac{1}{2}\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\Big[2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big]=\frac{\text{x}^2}{(\text{x}^2+\text{y}^2)}\bigg[\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\frac{\text{d}}{\text{dx}}(\text{x})}{\text{x}^2}\bigg]$
$ \Rightarrow\frac{1}{2}\Big(\frac{1}{\text{x}^2+\text{y}^2}\Big)\times2\Big(\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{x}^2}{(\text{x}^2+\text{y}^2)}\bigg[\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}(1)}{\text{x}^2}\bigg]$
$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}\frac{\text{dy}}{\text{dx}}=-\text{y}-\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\text{y}-\text{x})=-(\text{y}+\text{x})$

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