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If $y-x \sin y$, prove that $\frac{d y}{d x}=\frac{\sin y}{(1-x \cos y)}$


We are given with an equation $y=x \sin y$, we have to prove that $\frac{d y}{d x}=\frac{\sin y}{1-x \cos y}$ 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,

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

$\frac{d y}{d x}[1-x \cos y]=\sin y$

$\frac{d y}{d x}=\frac{\sin y}{1-x \cos y}$

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