#### explain solution RD Sharma class 12 chapter Differentiation exercise 10.4 question 8 maths

$\frac{y-4 x^{3}-4 x y^{2}}{4 y x^{2}+4 y^{3}-x}$

Hint:

Use chain rule and the product rule of differentiation

Given:

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

Solution:

Differentiate the given equation w.r.t x

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

$\frac{d}{d x}\left[\left(x^{2}\right)^{2}+\left(y^{2}\right)^{2}+2 x^{2} \times y^{2}\right]=x \frac{d y}{d x}+y \frac{d x}{d x} \quad\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$

[Product Rule  $\frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}$]

$\frac{d}{d x}\left[x^{4}+y^{4}+2 x^{2} y^{2}\right]=x \frac{d y}{d x}+y$

$\frac{d}{d x}\left(x^{4}\right)+\frac{d}{d x}\left(y^{4}\right)+\frac{d}{d x}\left(2 x^{2} y^{2}\right)=x \frac{d y}{d x}+y$

$4 x^{3}+\left(\frac{d y^{4}}{d y} \times \frac{d y}{d x}\right)+2\left(\frac{d\left(x^{2} y^{2}\right)}{d x}\right)=x \frac{d y}{d x}+y$

$4 x^{3}+4 y^{3} \frac{d y}{d x}+2\left[x^{2} \times \frac{d y^{2}}{d x}+y^{2} \frac{d x^{2}}{d x}\right]=x \frac{d y}{d x}+y \quad\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

$4 x^{3}+4 y^{3} \frac{d y}{d x}+2\left[x^{2} \times \frac{d y^{2}}{d y} \times \frac{d y}{d x}+y^{2}(2 x)\right]=x \frac{d y}{d x}+y$

$4 x^{3}+4 y^{3} \frac{d y}{d x}+2\left[x^{2} \times(2 y) \frac{d y}{d x}+2 x y^{2}\right]=x \frac{d y}{d x}+y$

$4 x^{3}+4 y^{3} \frac{d y}{d x}+4 x^{2} y \frac{d y}{d x}+4 x y^{2}=x \frac{d y}{d x}+y$

$4 y^{3} \frac{d y}{d x}+4 x^{2} y \frac{d y}{d x}-x \frac{d y}{d x}=y-4 x^{3}-4 x y^{2}$

$\frac{d y}{d x}=\frac{y-4 x^{3}-4 x y^{2}}{4 y x^{2}+4 y^{3}-x}$

Hence  $\frac{d y}{d x}=\frac{y-4 x^{3}-4 x y^{2}}{4 y x^{2}+4 y^{3}-x}$  is the required answer