#### need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 11

$\left[\frac{\sin (x+y)-y \cos x y}{x \cos x y-\sin (x+y)}\right]$

Hint:

Use chain rule and product rule

Given:

$\sin (x y)+\cos (x+y)=1$

Solution:

Differentiate the given equation w.r.t x

$\frac{d}{d x}[\sin x y+\cos (x+y)]=\frac{d(1)}{d x}$

$\frac{d}{d x}[\sin x y]+\frac{d}{d x}[\cos (x+y)]=0 \quad\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$

$\frac{d(\sin x y)}{d x y} \times \frac{d x y}{d x}+\frac{d \cos (x+y)}{d(x+y)} \times \frac{d(x+y)}{d x}=0$                    [Using chain rule]

$\cos (x y) \times\left[x \frac{d y}{d x}+y \frac{d x}{d x}\right]+(-\sin (x+y))\left[\frac{d x}{d x}+\frac{d y}{d x}\right]=0$            $\left[\begin{array}{c} \frac{d(\sin x)}{d x}=\cos x \\ \frac{d(\cos x)}{d x}=-\sin x \\ \frac{d(u v)}{d x}=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x} \end{array}\right]$

$\cos x y\left[x \frac{d y}{d x}+y\right]+(-\sin (x+y))\left[1+\frac{d y}{d x}\right]=0$

$x \cos x y \cdot \frac{d y}{d x}+y \cos x y-\sin (x+y)-\sin (x+y) \frac{d y}{d x}=0$

$\frac{d y}{d x}(x \cos x y-\sin (x+y))=\sin (x+y)-y \cos x y$

$\frac{d y}{d x}=\frac{\sin (x+y)-y \cos x y}{x \cos x y-\sin (x+y)}$

Hence, $\frac{d y}{d x}=\frac{\sin (x+y)-y \cos x y}{x \cos x y-\sin (x+y)}$ is the required answer.