Question
If $\text{y}=\sqrt{\text{x}^2+\text{a}^2},$ prvoe that $\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}=0$

Answer

Here, $\text{y}=\sqrt{\text{x}^2+\text{a}^2}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}^2+\text{a}^2}\big)$
$=\frac{1}{2\sqrt{\text{x}^2+\text{a}^2}}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{a}^2\big)$
[Using chain rule]
$=\frac{1}{2\sqrt{\text{x}^2+\text{a}^2}}\times(2\text{x})$
$=\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\ \Big[\text{Since},\sqrt{\text{x}^2+\text{a}^2}=\text{y}\Big]$
$\Rightarrow \text{y}\frac{\text{dy}}{\text{dx}}=\text{x}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}=0$

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