If xy=1, prove that dy dx +y^2=0 - Vidyadip

Question

If $\text{xy}=1,$ prove that $\frac{\text{dy}}{\text{dx}}+\text{y}^2=0$

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Answer

Here, xy = 1 .....(i)
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}(\text{xy})=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})=0$
[Using product rule]
$ \Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)=0$
$\Big[\text{Put x}=\frac{1}{\text{y}}\text{ from equation (i)}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\frac{1}{\text{y}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\text{y}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\text{y}^2=0$

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