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

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


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


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


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]

Is the required answer.

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



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