1. **State the problem:** Find the $n^{th}$ order derivative of the function $$y = \sin(5x) \cdot \cos(3x).$$ 2. **Use product rule:** Since $y$ is a product of two functions, $u = \sin(5x)$ and $v = \cos(3x)$, the derivative is $$y' = u'v + uv'.$$ 3. **Find first derivatives:** $$u' = \frac{d}{dx} \sin(5x) = 5 \cos(5x),$$ $$v' = \frac{d}{dx} \cos(3x) = -3 \sin(3x).$$ 4. **First derivative:** $$y' = 5 \cos(5x) \cdot \cos(3x) - 3 \sin(5x) \cdot \sin(3x).$$ 5. **Use trigonometric identity:** Recall $$\cos A \cos B - \sin A \sin B = \cos(A + B).$$ So, $$y' = 5 \cos(5x) \cos(3x) - 3 \sin(5x) \sin(3x) = 5 \cos(8x) - 3 \cos(2x).$$ 6. **Rewrite $y$ using product-to-sum:** $$y = \sin(5x) \cos(3x) = \frac{1}{2} [\sin(8x) + \sin(2x)].$$ 7. **General $n^{th}$ derivative:** The $n^{th}$ derivative of $\sin(kx)$ is $$\frac{d^n}{dx^n} \sin(kx) = k^n \sin\left(kx + n \frac{\pi}{2}\right).$$ 8. **Apply to $y$:** $$y^{(n)} = \frac{1}{2} \left[8^n \sin\left(8x + n \frac{\pi}{2}\right) + 2^n \sin\left(2x + n \frac{\pi}{2}\right)\right].$$ **Final answer:** $$\boxed{y^{(n)} = \frac{1}{2} \left[8^n \sin\left(8x + n \frac{\pi}{2}\right) + 2^n \sin\left(2x + n \frac{\pi}{2}\right)\right]}.$$