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

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