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



$y=x \sin y$

siny $=\frac{y}{x}$     ....(1)

Differentiating it with respect to $x$ using product rule,

$\frac{d y}{d x}=\frac{d}{d x}(x$ siny $)$

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

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

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

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

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

$\frac{d y}{d x}=\frac{y}{x(1-x \cos y)}[$ From (i) $]$

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