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Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Multiple choice question 19

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Answer:

        \frac{\sin ^{2}(a+y)}{\sin a}

Hint:

        Differentiate the function w.r.t x

Given:

        \sin y=x \sin (a+y), \text { then } \frac{d y}{\alpha x}

Solution:  

        \begin{aligned} &\sin y=x \sin (a+y) \\\\ &\frac{d}{d x}(\sin y)=\frac{d}{d x}[(a+y) \sin x] \end{aligned}

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

                        \begin{aligned} &=\sin (a+y) \times 1+x \cos (a+y) \frac{d y}{d x} \\ &=\sin (a+y)+x \cos (a+y) \frac{d y}{d x} \end{aligned}

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

        \left(\cos y-\frac{\sin y}{\sin (a+y)} \times \cos (a+y)\right) \frac{d y}{d x}=\sin (a+y)                \left[\begin{array}{l} \sin y=2 \sin x \cos x \\\\ x=\frac{\sin y}{\sin (a+y)} \end{array}\right]

        \left(\frac{\sin (a+y) \cos y-\sin y \cos (a+y)}{\sin (a+y)}\right) \frac{d y}{d x}=\sin (a+y)

        \begin{aligned} &\frac{\sin (a+y-y)}{\sin (a+y)} \times \frac{d y}{d x}=\sin (a+y) \\\\ &\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a} \end{aligned}

 

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