Question

find dy/dx of (x^2+y^2)^2=xy

Answer 1

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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 with respect to x using the chain rule, we get

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

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

$4x^3+4x^2y\frac{dy}{dx}+4xy^2+4y^3\frac{dy}{dx}=y+x\frac{dy}{dx}$

$4x^3-y=x\frac{dy}{dx}-4xy^2\frac{dy}{dx}-4x^2y\frac{dy}{dx}$

$4x^3-y=\frac{dy}{dx}(x-4x^y-4xy^2)$

$\frac{dy}{dx}=\frac{4x^3-y}{x-4x^2y-4xy^2}$.