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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
\[\text{ We have, y } = x \sin\left( a + y \right) \]
Differentiate with respect to y,
\[\frac{d y}{d x} = \frac{d}{dx}\left[ x \sin\left( a + y \right) \right]\]
\[ \Rightarrow \frac{d y}{d x} = x\frac{d}{dx}\left\{ \sin\left( a + y \right) \right\} + \sin\left( a + y \right)\frac{d}{dx}\left( x \right) ..........\left[\text{using product rule and chain rule} \right]\]
\[ \Rightarrow \frac{d y}{d x} = x \cos\left( a + y \right)\frac{d}{dx}\left( a + y \right) + \sin\left( a + y \right)\left( 1 \right)\]
\[ \Rightarrow \frac{d y}{d x}\left\{ 1 - x \cos\left( a + y \right) \right\} = \sin\left( a + y \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{\sin\left( a + y \right)}{1 - x \cos\left( a + y \right)}\]
\[ \Rightarrow \frac{d y}{d x} = \frac{\sin\left( a + y \right)}{1 - \frac{y}{\sin\left( a + y \right)} \cos\left( a + y \right)}\left[ \because y = x\sin\left( a + y \right) \right]\]
\[ \Rightarrow \frac{d y}{d x} = \frac{\sin^2 \left( a + y \right)}{\sin \left( a + y \right) - y \cos \left( a + y \right)}\]
Hence proved
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