1. **State the problem:** We are given the function $f(x) = 4 \sin x + 7 \cos x$ and need to find its derivative $f'(x)$ and then evaluate $f'(x)$ at $x = \frac{5\pi}{6}$. 2. **Recall the derivative rules:** - The derivative of $\sin x$ is $\cos x$. - The derivative of $\cos x$ is $-\sin x$. 3. **Find the derivative $f'(x)$:** $$ f'(x) = 4 \frac{d}{dx}(\sin x) + 7 \frac{d}{dx}(\cos x) = 4 \cos x - 7 \sin x $$ 4. **Evaluate $f'(x)$ at $x = \frac{5\pi}{6}$:** - Calculate $\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$ - Calculate $\sin \frac{5\pi}{6} = \frac{1}{2}$ Substitute these values: $$ f'\left(\frac{5\pi}{6}\right) = 4 \left(-\frac{\sqrt{3}}{2}\right) - 7 \left(\frac{1}{2}\right) = -2\sqrt{3} - \frac{7}{2} $$ 5. **Final answer:** $$ f'(x) = 4 \cos x - 7 \sin x $$ $$ f'\left(\frac{5\pi}{6}\right) = -2\sqrt{3} - \frac{7}{2} $$