1. Problem: Find the derivative of $y = \sin \alpha x$. 2. Formula: The derivative of $\sin u$ with respect to $x$ is $\cos u \cdot \frac{du}{dx}$. 3. Applying the chain rule: Here, $u = \alpha x$, so $\frac{du}{dx} = \alpha$. 4. Therefore, $$y' = \cos \alpha x \cdot \alpha = \alpha \cos \alpha x.$$