1. **State the problem:** We are given the function $$f(x) = 6 + x^2 + \sin 4x$$ and need to find its derivative $$f'(x)$$ with respect to $$x$$. 2. **Recall the derivative rules:** - The derivative of a constant is 0. - The derivative of $$x^n$$ is $$nx^{n-1}$$. - The derivative of $$\sin u$$ where $$u$$ is a function of $$x$$ is $$\cos u \cdot u'$$ (chain rule). 3. **Apply the derivative to each term:** - Derivative of 6 is 0. - Derivative of $$x^2$$ is $$2x$$. - Derivative of $$\sin 4x$$ is $$\cos 4x \cdot \frac{d}{dx}(4x) = \cos 4x \cdot 4$$. 4. **Combine the results:** $$f'(x) = 0 + 2x + 4\cos 4x = 2x + 4\cos 4x$$. 5. **Final answer:** $$f'(x) = 2x + 4\cos 4x$$