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let ( f(x)=7cos x - 7\tan x ). ( f(x)=) ( f(\frac{pi}{3})=)

Question

let ( f(x)=7cos x - 7\tan x ).
( f(x)=)
( f(\frac{pi}{3})=)

Explanation:

Step1: Differentiate \( f(x) \)

Use the sum rule \((u - v)^\prime=u^\prime - v^\prime\) and the derivatives \((\cos x)^\prime=-\sin x\) and \((\tan x)^\prime=\sec^{2}x\).
If \(f(x)=7\cos x - 7\tan x\), then \(f^\prime(x)=7(\cos x)^\prime-7(\tan x)^\prime\).
\(f^\prime(x)=-7\sin x - 7\sec^{2}x\).

Step2: Evaluate \(f^\prime(\frac{\pi}{3})\)

Recall that \(\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\) and \(\sec(\frac{\pi}{3}) = 2\) (since \(\sec x=\frac{1}{\cos x}\) and \(\cos(\frac{\pi}{3})=\frac{1}{2}\)).
Substitute \(x = \frac{\pi}{3}\) into \(f^\prime(x)\):
\(f^\prime(\frac{\pi}{3})=-7\times\frac{\sqrt{3}}{2}-7\times2^{2}\).
\(f^\prime(\frac{\pi}{3})=-\frac{7\sqrt{3}}{2}-28\).

Answer:

\(f^\prime(x)=-7\sin x - 7\sec^{2}x\); \(f^\prime(\frac{\pi}{3})=-\frac{7\sqrt{3}}{2}-28\)