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

$\frac{\sin ^{2}(a+y)}{\sin a}$

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}