Question

${\left(x^2+y^2\right)}^2=xy,$ then $\frac{dy}{dx}$

Answer 1

We have,

$(x^2+y^2{)}^2=xy$

On differentiating w.r.t $x$, we get

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

$2(x^2+y^2)\frac{d}{dx}(x^2+y^2)=x\frac{d}{dx}\left(y\right)+y\frac{d}{dx}\left(x\right)$

$2(x^2+y^2)(2x+2y\frac{dy}{dx})=x\frac{dy}{dx}+y\times 1$

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

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

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

$\left[4xy\right(x^2+y^2)-x]\frac{dy}{dx}=y-4x(x^2+y^2)$

$\frac{dy}{dx}=\frac{y-4x(x^2+y^2)}{\left[4xy\right(x^2+y^2)-x]}$

Hence, this is the answer.