$y=(\sin x+\cos x)^x$, then $\frac{d y}{d x}$ is

y $\left\{\frac{\log y}{x}+\frac{x(\cos x-\sin x)}{\sin x+\cos x}\right\}$

$y=(\sin x+\cos x)^{x}$

$\log y=x \log (\sin x+\cos x)$

$\frac{1}{y} . \frac{d y}{d x}=\log (\sin x+\cos x)+\frac{x \times(\cos x-\sin x)}{\sin x+\cos x}$

$\frac{d y}{d x}=(\sin x+\cos x)^{x}\left[\frac{x(\cos x-\sin x}{\sin x+\cos x}+\log (\sin x+\cos x)\right] = y \left\{\frac{\log y}{x}+\frac{x(\cos x-\sin x)}{\sin x+\cos x}\right\}$

Hence (2) is correct answer.