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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,

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

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

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