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

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

Answers (1)

Answer:

\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.

 

 

Posted by

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