let\n$f(x)=3\\cos x - 6\\tan x$\n$f(x)=$\n$f(\\frac{\\pi}{3})=$\nquestion help: video message instruc

Answer:

$f'(x)=-3\sin x - 6\sec^{2}x$; $f'(\frac{\pi}{3})=-3\sqrt{3}-24$

Explanation:

Step1: Differentiate each term

• Recall the derivative of $\cos x$ is $-\sin x$ and the derivative of $\tan x$ is $\sec^{2}x$.
• For $y = 3\cos x$, using the constant - multiple rule $(cf(x))'=cf'(x)$ where $c = 3$ and $f(x)=\cos x$, we have $(3\cos x)'=3\times(-\sin x)=-3\sin x$.
• For $y=-6\tan x$, using the constant - multiple rule with $c = - 6$ and $f(x)=\tan x$, we get $(-6\tan x)'=-6\times\sec^{2}x$.

Step2: Find $f'(x)$

• By the sum rule $(u + v)'=u'+v'$ where $u = 3\cos x$ and $v=-6\tan x$, $f'(x)=(3\cos x - 6\tan x)'=(3\cos x)'+(-6\tan x)'=-3\sin x-6\sec^{2}x$.

Step3: Evaluate $f'(\frac{\pi}{3})$

• Substitute $x = \frac{\pi}{3}$ into $f'(x)$.
• We know that $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$ and $\sec\frac{\pi}{3}=\frac{1}{\cos\frac{\pi}{3}} = 2$.
• $f'(\frac{\pi}{3})=-3\times\frac{\sqrt{3}}{2}-6\times2^{2}$.
• $f'(\frac{\pi}{3})=-\frac{3\sqrt{3}}{2}-24=-3\sqrt{3}-24$.