91Ó°ÊÓ

By applying the product rule, we have:$$y_{1}^{\prime} = y^{\prime} = \frac{d}{dx} \left(\frac{1}{2} e^{x} \cos x\right)$$First, let's assign \(f(x) = \frac{1}{2} e^{x}\) and \(g(x) = \cos x\). Then, find their first derivatives:$$f^{\prime}(x) = \frac{d}{dx} \left(\frac{1}{2} e^{x}\right) = \frac{1}{2} e^{x}$$$$g^{\prime}(x) = \frac{d}{dx} (\cos x) = -\sin x$$Now, apply the product rule:$$y^{\prime} = f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x) = \frac{1}{2} e^{x} \cdot \cos x + \frac{1}{2} e^{x} \cdot -\sin x$$Thus, the first derivative is:$$y^{\prime} = \frac{1}{2} e^{x} (\cos x - \sin x)$$

Again, apply the product rule to find the second derivative, \(y^{\prime\prime}\), by differentiating \(y^{\prime}\). Let \(h(x)=\frac{1}{2} e^{x}\) and \(p(x)=\cos x - \sin x\). Then, find their first derivatives:$$h^{\prime}(x) = \frac{d}{dx} \left(\frac{1}{2} e^{x}\right) = \frac{1}{2} e^{x}$$$$p^{\prime}(x) = \frac{d}{dx} (\cos x - \sin x)=-\sin x - \cos x$$Now, apply the product rule:$$y^{\prime\prime} = h^{\prime}(x) \cdot p(x) + h(x) \cdot p^{\prime}(x)= \frac{1}{2} e^{x} \cdot (\cos x - \sin x) + \frac{1}{2} e^{x} \cdot (-\sin x - \cos x)$$Thus, the second derivative is:$$y^{\prime\prime} = \frac{1}{2} e^{x} (\cos x - \sin x - \sin x - \cos x) = \frac{1}{2} e^{x} (-2 \sin x)$$Therefore, the second derivative \(y^{\prime\prime}\) of the given function is:$$y^{\prime\prime} = -e^{x} \sin x$$