Question

If $(x^2+y^2{)}^2=xy$, find $\frac{dy}{dx}$.

Answer 1

Now the given equation is $(x^2+y^2{)}^2=xy$

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

$\to x^4+xy+y^4=0$

Differentiating the above expression wrt $x$, we get

$4x^3+x\frac{dy}{dx}+y+4y^3\frac{dy}{dx}=0$

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

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