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Question
If \[y = x \sin y\] , Prove that \[\frac{dy}{dx} = \frac{\sin y}{\left( 1 - x \cos y \right)}\] ?
Solution
\[\text{ We have, y = x } \sin y\]
Differentiating with respect to x, we get
\[\frac{d y}{d x} = \frac{d}{dx}\left( x \sin y \right)\]
\[ \Rightarrow \frac{d y}{d x} = x\frac{d}{dx}\left( \sin y \right) + \sin y\frac{d}{dx}\left( x \right)\]
\[ \Rightarrow \frac{d y}{d x} = x\cos y\frac{d y}{d x} + \sin y\]
\[ \Rightarrow \frac{d y}{d x}\left( 1 - x\cos y \right) = \sin y\]
\[ \Rightarrow \frac{d y}{d x} = \frac{\sin y}{1 - x\cos y}\]
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