compute the following for (f(x)=3sin x + 3cos x).\na. (f(x)=)\nb. (f(\frac{2pi}{3})=)

Explanation:

Step1: Apply derivative rules

The derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin x$. Using the sum - rule of derivatives $(u + v)'=u'+v'$ for $u = 3\sin x$ and $v = 3\cos x$. So $f'(x)=\frac{d}{dx}(3\sin x)+\frac{d}{dx}(3\cos x)=3\cos x- 3\sin x$.

Step2: Evaluate $f'(\frac{2\pi}{3})$

Substitute $x = \frac{2\pi}{3}$ into $f'(x)$. We know that $\cos(\frac{2\pi}{3})=-\frac{1}{2}$ and $\sin(\frac{2\pi}{3})=\frac{\sqrt{3}}{2}$. Then $f'(\frac{2\pi}{3})=3\cos(\frac{2\pi}{3})-3\sin(\frac{2\pi}{3})=3\times(-\frac{1}{2})-3\times\frac{\sqrt{3}}{2}=-\frac{3 + 3\sqrt{3}}{2}$.

Answer:

a. $3\cos x-3\sin x$ b. $-\frac{3 + 3\sqrt{3}}{2}$