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

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

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Answer:

\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

 

Posted by

infoexpert26

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