Question
If $\text{y}=\frac{\text{x}}{\text{x}+2},$ show that $\text{x}\frac{\text{dy}}{\text{dx}}=(1-\text{y})\text{y}$
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Answer
We have, $\text{y}=\frac{\text{x}}{\text{x}+2}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{x}+2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}+2)\frac{\text{d}}{\text{dx}}(\text{x})-\text{x}\frac{\text{d}}{\text{dx}}(\text{x}+2)}{(\text{x}+2)^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2-\text{x}}{(\text{x}+2)^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2}{(\text{x}+2)^2}-\frac{\text{x}}{(\text{x}+2)^2}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}+2}-\frac{\text{xy}^2}{\text{x}^2} \Big[\because\ \text{x}+2=\frac{\text{x}}{\text{y}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\frac{\text{y}^2}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\text{y}(1-\text{y})$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=(1-\text{y})\text{y}$
Hence, proved.
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{x}+2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}+2)\frac{\text{d}}{\text{dx}}(\text{x})-\text{x}\frac{\text{d}}{\text{dx}}(\text{x}+2)}{(\text{x}+2)^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2-\text{x}}{(\text{x}+2)^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2}{(\text{x}+2)^2}-\frac{\text{x}}{(\text{x}+2)^2}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}+2}-\frac{\text{xy}^2}{\text{x}^2} \Big[\because\ \text{x}+2=\frac{\text{x}}{\text{y}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\frac{\text{y}^2}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\text{y}(1-\text{y})$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=(1-\text{y})\text{y}$
Hence, proved.
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