Question
यदि $\left(x^2+y^2\right)^2=x y,$ तब $\frac{d y}{d x}$ है $-$

Answer

$(C)\ \frac{y-4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}$
यहाँ $\left(x^2+y^2\right)^2=x y$
$x$ के सापेक्ष अवकलन करने पर,
$ 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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