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Question
If \[y = x \sin \left( a + y \right)\] , prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin \left( a + y \right) - y \cos \left( a + y \right)}\] ?
Solution
\[\text{ We have, y } = x \sin\left( a + y \right)\]
Differentiating with respect to x using chain rule,
\[\frac{dy}{dx} = x\frac{d}{dx}\sin\left( a + y \right) + \sin\left( a + y \right)\frac{d}{dx}\left( x \right) \]
\[ \Rightarrow \frac{dy}{dx} = x \cos\left( a + y \right)\frac{dy}{dx} + \sin\left( a + y \right)\]
\[ \Rightarrow \left\{ 1 - x \cos\left( a + y \right) \right\}\frac{dy}{dx} = \sin\left( a + y \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin\left( a + y \right)}{1 - x \cos\left( a + y \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin\left( a + y \right)}{1 - \frac{y}{\sin\left( a + y \right)}\cos\left( a + y \right)} .............\left[ \because \frac{y}{\sin\left( a + y \right)} = x \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin\left( a + y \right) - y \cos\left( a + y \right)}\]
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