If \(\sin \left(\frac{x+y}{x-y}\right)=\tan \frac{\pi}{5}\), then \(\frac{d y}{d x}=\)

Given \(\sin \left(\frac{x+y}{x-y}\right)=\tan \frac{\pi}{5}\)
\(\therefore \frac{x+y}{x-y}=\sin ^{-1}\left(\tan \frac{\pi}{5}\right) \Rightarrow \frac{x+y}{x-y}=K \quad \ldots\) say ...(1)
\(\therefore x+y=K(x-y)\)
\(\quad 1+\frac{d y}{d x}=K\left(1-\frac{d y}{d x}\right) \Rightarrow 1+\frac{d y}{d x}=K-K \frac{d y}{d x}\)
\((1+K) \frac{d y}{d x}=K-1 \Rightarrow \frac{d y}{d x}=\frac{K-1}{K+1}\)
\(\begin{aligned} \frac{d y}{d x} &=\frac{\frac{x+y}{x-y}-1}{\frac{x+y}{x-y}+1} \quad[\ldots \text { From (1) }] \\ &=\frac{x+y-x+y}{x+y+x-y}=\frac{2 y}{2 x}=\frac{y}{x} \end{aligned}\)
This problem can also be solved as follow:
\(\frac{x+y}{x-y}=\sin ^{-1}\left(\tan \frac{\pi}{5}\right)\)
Differentiating w.r.t. \(\mathrm{x}\), we get
\(\begin{aligned}
& \frac{(x-y)\left(1+\frac{d y}{d x}\right)-(x+y)\left(1-\frac{d y}{d x}\right)}{(x-y)^{2}}=0 \\
\therefore &\left(x-y+x \frac{d y}{d x}-y \frac{d y}{d x}\right)-\left(x+y-x \frac{d y}{d x}-y \frac{d y}{d x}\right)=0 \\
\therefore &-2 y+2 x \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{y}{x}
\end{aligned}\)