Question

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

Answer 1

We have,

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

On differentiating both sides with respect to $x$, we have

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

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

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

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

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

Hence, this is the answer.