Question
If $\left(x^2+y^2\right)^2=x y$, then $\frac{d y}{d x}$ :

Answer

$(C)$ Here $\left(x^2+y^2\right)^2=x y$
Differentiating $w.r.t. x^ 2\left(x^2+y^2\right)\left(2 x+2 y \frac{d y}{d x}\right)=x \frac{d y}{d x}+y$
$\Rightarrow 4\left(x^2+y^2\right) x+4\left(x^2+y^2\right) y \frac{d y}{d x}=x \frac{d y}{d x}+y$
$\Rightarrow \quad\left[4 y\left(x^2+y^2\right)-x\right] \frac{d y}{d x}=y-4 x\left(x^2+y^2\right)$
Hence, $\frac{d y}{d x}=\frac{y-4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}$

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