Find dy dx in the following cases: (x^2 + y^2)^2 = xy - Vidyadip

Question

Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$(x^2 + y^2)^2 = xy$

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Answer

Given, $(x^2 + y^2)^2 = xy$
Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}\Big(\big(\text{x}^2+\text{y}^2\big)^2\Big)=\frac{\text{d}}{\text{dx}}(\text{xy})$
$\Rightarrow2(\text{x}^2+\text{y}^2\big)\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y}^2\big)=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}\big(\text{x}\big)$
[Using chain rule]
$\Rightarrow2(\text{x}^2+\text{y}^2\big)\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)$
$\Rightarrow4\text{x}(\text{x}^2+\text{y}^2\big)+4\text{y}\big(\text{x}^2+\text{y}^2\big)\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}$
$\Rightarrow4\text{y}\big(\text{x}^2+\text{y}^2\big)\frac{\text{dy}}{\text{dx}}-\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-4\text{x}(\text{x}^2+\text{y}^2\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[4\text{y}\text{x}^2+4\text{y}^3-\text{x}\big]=\text{y}-4\text{x}^3-4\text{x}\text{y}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{y}-4\text{x}^3-4\text{x}\text{y}^2}{4\text{y}\text{x}^2+4\text{y}^3-\text{x}}\Big)$

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