Question

If \[y = x \sin y\] , prove that  \[\frac{dy}{dx} = \frac{y}{x \left( 1 - x \cos y \right)}\] ?

 

Answer 1

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

Differentiating with respect to x,

\[\frac{dy}{dx} = \frac{d}{dx}\left( x \sin y \right)\]
\[ \Rightarrow \frac{dy}{dx} = x\frac{d}{dx}\left( \sin y \right) + \sin y\frac{d}{dx}\left( x \right) \]
\[ \Rightarrow \frac{dy}{dx} = x \cos y\frac{dy}{dx} + \sin y\left( 1 \right)\]
\[ \Rightarrow \frac{dy}{dx} - x \cos y\frac{dy}{dx} = \sin y\]
\[ \Rightarrow \frac{dy}{dx}\left( 1 - x \cos y \right) = \sin y\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin y}{\left( 1 - x \cos y \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y}{x\left( 1 - x \cos y \right)} \left[ \because \sin y = \frac{y}{x} \right]\]