provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 2

$\frac{(x+y)^{2}}{y^{2}-2 x y-x^{2}}$

Hint:

Use product rule to find $\frac{d y}{d x}$

Given:

$y^{3}-3 x y^{2}=x^{3}+3 x^{2} y$

Solution:

Differentiating the given equation w.r.t x

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

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

$3 y^{2} \frac{d y}{d x}-3\left[x \frac{d y^{2}}{d x}+y^{2} \frac{d x}{d x}\right]=3 x^{2}+3\left[x^{2} \frac{d y}{d x}+y \frac{d x^{2}}{d x}\right]$                $\left[\because \frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$

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

$3 y^{2} \frac{d y}{d x}-3\left[x(2 y) \cdot \frac{d y}{d x}+y^{2}\right]=3 x^{2}+3 x^{2} \frac{d y}{d x}+6 x y$                        $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

$3 y^{2} \frac{d y}{d x}-6 x y \frac{d y}{d x}-3 y^{2}=3 x^{2}+3 x^{2} \frac{d y}{d x}+6 x y$

$3 y^{2} \frac{d y}{d x}-6 x y \frac{d y}{d x}-3 x^{2} \frac{d y}{d x}=3 x^{2}+6 x y+3 y^{2}$

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

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

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

Hence $\frac{d y}{d x}=\frac{(x+y)^{2}}{y^{2}-2 x y-x^{2}}$