Question
If $\text{y}=\sqrt{\log\text{x}+\sqrt{\log\text{x}+\sqrt{\log\text{x}+\ .... \text{to }\infty}}},$ prove that $(2\text{y}-1)\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$

Answer

We have, $\text{y}=\sqrt{\log\text{x}+\sqrt{\log\text{x}+\sqrt{\log\text{x}+\ .... \text{to }\infty}}}$
$\Rightarrow\text{y}=\sqrt{\log\text{x}+\text{y}}$
Squaring both sides, we get,
$\text{y}^2=\log\text{x}+\text{y}$
$=2\text{y}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}+\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text{y}-1)=\frac{1}{\text{x}}$

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