Question:
If $y=x \sin (a+y)$, prove that $\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin (a+y)-y \cos (a+y)}$
Solution:
We are given with an equation $y=x \sin (a+y)$, we have to prove that $\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin (a+y)-y \cos (a+y)}$ by using the
given equation we will first find the value of $\frac{d y}{d x}$ and we will put this in the equation we have to prove, so by differentiating the equation on both sides with respect to $x$, we get,
$\frac{d y}{d x}=(1) \sin (a+y)+x \cos (a+y) \frac{d y}{d x}$
$\frac{d y}{d x}=\frac{\sin (a+y)}{1-x \cos (a+y)}$
We can further solve it by using the given equation,
$\frac{d y}{d x}=\frac{\sin (a+y)}{1-\frac{y}{\sin (a+y)} \cos (a+y)}$
$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin (a+y)-y \cos (a+y)}$