#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 25 maths textbook solution

$\frac{d y}{d x}=\frac{2 x^{3}+y-x^{2} y \cos x y}{x\left(x^{2} \cos x y+1+2 x y\right)}$

Hint:

Use chain rule and $\frac{d\left(x^{n}\right)}{d x}=n x^{n-1}$

Given:

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

Solution:

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

Differentiate w.r.t x

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

$\frac{d(\sin x y)}{d(x y)} \times \frac{d(x y)}{d x}+\frac{d}{d x}\left(\frac{y}{x}\right)=2 x-\frac{d y^{2}}{d y} \times \frac{d y}{d x}$                        [Using chain rule and $\frac{d\left(x^{n}\right)}{d x}=n x^{n-1}$]

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

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

$x \cos x y \frac{d y}{d x}+y \cos x y+\frac{1}{x} \frac{d y}{d x}-\frac{y}{x^{2}}=2 x-2 y \frac{d y}{d x}$

$x \cos x y \frac{d y}{d x}+\frac{1}{x} \frac{d y}{d x}+2 y \frac{d y}{d x}=2 x+\frac{y}{x^{2}}-y \cos x y$

$\frac{d y}{d x}\left(x \cos x y+\frac{1}{x}+2 y\right)=2 x+\frac{y}{x^{2}}-y \cos x y$

$\frac{d y}{d x}\left(\frac{x^{2} \cos x y+1+2 x y}{x}\right)=\frac{2 x^{3}+y-x^{2} y \cos x y}{x^{2}}$

$\frac{d y}{d x}\left(x^{2} \cos x y+2 x y+1\right)=\frac{1}{x}\left(2 x^{3}+y-x^{2} y \cos x y\right)$

$\frac{d y}{d x}=\frac{2 x^{3}+y-x^{2} y \cos x y}{x\left(x^{2} \cos x y+1+2 x y\right)}$

Hence proved