Question

(x square +y square)whole square=xy , find dy by dx

Answer 1

${x}^{2} \ + {y}^{2} = xy$
differenting both sides
2x +2yy'=y+ xy'
2x-y=y'(x-2y)
2x-y/(x-2y)=y'
This is your answer
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Answer 2

Answer:

Step-by-step explanation:

The given equation is $(x^{2}+y^{2})^{2}=xy$.

Differentiating the above equation, we get

$2(x^{2}+y^{2})^{2}(2x+2y\frac{dy}{dx})= x\frac{dy}{dx}+y$

$4x(x^{2}+y^{2})^{2}+4y(x^{2}+y^{2})^{2}\frac{dy}{dx}=x\frac{dy}{dx}+y$

$4x(x^{2}+y^{2})^{2}-y=(x-4y(x^{2}+y^{2})^{2})\frac{dy}{dx}$

$\frac{4x(x^{2}+y^{2})^{2}-y}{x-4y(x^{2}+y^{2})^{2}}=\frac{dy}{dx}$

which is the required solution.