Question

If \[y = x \sin \left( a + y \right)\] ,Prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin \left( a + y \right) - y \cos \left( a + y \right)}\] ?

Answer 1

\[\text{ We have, y }   = x \sin\left( a + y \right) \]

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