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

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


Use chain rule and product rule


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


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.



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