Example 5 Find the extremum of $y=e^{\prime} \sin x$.

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However, it seems there is a typographical error in the original function, as $e^{\prime}$ is not a standard notation. The correct function might be $y=e \sin x$ or $y=e^x \sin x$. Could you please clarify which function you are referring to?
Solve $y^{\prime}=e^{x}(\sin x+\cos x)$, let $y^{\prime}=0$ to get $x=2 k \pi-\frac{\pi}{4}$ or $2 k \pi+\frac{3}{4} \pi, k=0, \pm 1, \cdots$
Since $y^{\prime \prime}=2 e^{x} \cos x,\left.y^{\prime \prime}\right|_{x=-\frac{\pi}{4}+2 k \pi}>0,\left.y^{\prime \prime}\right|_{x=\frac{3}{4} \pi+2 k \pi}<0$,
Therefore, when $x=-\frac{\pi}{4}+2 k \pi$ there is a minimum value $y=-\frac{\sqrt{2}}{2} e^{-\frac{\pi}{4}+2 k \pi}$; when $x=\frac{3}{4} \pi+2 k \pi$ there is a maximum value $y=\frac{\sqrt{2}}{2} e^{\frac{3}{1} \pi+2 k \pi}$, where $k \in \mathbf{Z}$.