#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 21 maths textbook solution

$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin (a+y)-y \cos (a+y)}$

Hint:

Use chain rule

Given:

$y=x \sin (a+y)$

Solution:

$y=x \sin (a+y)$

Differentiate the given equation w.r.t x

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

$\frac{d y}{d x}=x \cdot \frac{d \sin (a+y)}{d x}+\sin (a+y) \cdot \frac{d x}{d x}$

$\frac{d y}{d x}=x\left[\frac{d \sin (a+y)}{d(a+y)} \times \frac{d(a+y)}{d y} \times \frac{d y}{d x}\right]+\sin (a+y)$                    [Using chain rule]

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

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

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

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

$\frac{d y}{d x}(1-x \cos (a+y))=\sin (a+y)$

$\frac{d y}{d x}=\frac{\sin (a+y)}{1-x \cos (a+y)}$

$y=x \sin (a+y)$

$x=\frac{y}{\sin (a+y)}$

Put $x=\frac{y}{\sin (a+y)}$  in the above equation

$\frac{d y}{d x}=\frac{\sin (a+y)}{1-\frac{y}{\sin (a+y)} \times \cos (a+y)}$

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

$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin (a+y)-y \cos (a+y)}$

Thus, proved