1. The problem is to differentiate the product of two terms: $4 - x$ and $\sin\left(\frac{n \pi x}{4}\right)$.\n\n2. We treat $4 - x$ as one function, say $u = 4 - x$, and $\sin\left(\frac{n \pi x}{4}\right)$ as another function, say $v = \sin\left(\frac{n \pi x}{4}\right)$.\n\n3. To differentiate the product $uv$, we use the product rule: $\frac{d}{dx}(uv) = u'v + uv'$.\n\n4. First, find $u' = \frac{d}{dx}(4 - x) = -1$.\n\n5. Next, find $v' = \frac{d}{dx} \sin\left(\frac{n \pi x}{4}\right)$. Using the chain rule, $v' = \cos\left(\frac{n \pi x}{4}\right) \cdot \frac{d}{dx}\left(\frac{n \pi x}{4}\right) = \cos\left(\frac{n \pi x}{4}\right) \cdot \frac{n \pi}{4}$.\n\n6. Substitute $u$, $u'$, $v$, and $v'$ into the product rule formula:\n$$\frac{d}{dx} \left[(4 - x) \sin\left(\frac{n \pi x}{4}\right)\right] = (-1) \cdot \sin\left(\frac{n \pi x}{4}\right) + (4 - x) \cdot \cos\left(\frac{n \pi x}{4}\right) \cdot \frac{n \pi}{4}.$$\n\n7. Simplify the expression if desired:\n$$= -\sin\left(\frac{n \pi x}{4}\right) + \frac{n \pi}{4} (4 - x) \cos\left(\frac{n \pi x}{4}\right).$$\n\nThis is the derivative of the product $ (4 - x) \sin\left(\frac{n \pi x}{4}\right)$.