1. **State the problem:** Find the derivative of the function $g(x) = 3e^x \cos x$. 2. **Formula used:** To differentiate a product of two functions, use the product rule: $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$ 3. **Identify parts:** Here, $u(x) = 3e^x$ and $v(x) = \cos x$. 4. **Differentiate each part:** - $u'(x) = 3e^x$ because the derivative of $e^x$ is $e^x$ and constants stay. - $v'(x) = -\sin x$ because the derivative of $\cos x$ is $-\sin x$. 5. **Apply product rule:** $$g'(x) = u'(x)v(x) + u(x)v'(x) = 3e^x \cos x + 3e^x (-\sin x)$$ 6. **Simplify:** $$g'(x) = 3e^x \cos x - 3e^x \sin x = 3e^x (\cos x - \sin x)$$ **Final answer:** $$g'(x) = 3e^x (\cos x - \sin x)$$