Question

3.3b derivatives of trigonometric functions
(1 point)
let
f(x)=\frac{7 \tan (x)-12}{sec (x)}
find:
f^{prime}(x)=
f^{prime}left(\frac{11 pi}{6}
ight)=
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Explanation:

Step1: Simplify the function

Using the trigonometric identities \(\tan(x)=\frac{\sin(x)}{\cos(x)}\) and \(\sec(x)=\frac{1}{\cos(x)}\), we can rewrite \(f(x)\) as \(f(x)=7\sin(x)-12\cos(x)\).

Step2: Differentiate the function

The derivative of \(\sin(x)\) is \(\cos(x)\) and the derivative of \(\cos(x)\) is \(-\sin(x)\). Using the sum - rule of differentiation \((u + v)^\prime=u^\prime+v^\prime\) (where \(u = 7\sin(x)\) and \(v=-12\cos(x)\)), we get \(f^\prime(x)=7\cos(x)+12\sin(x)\).

Step3: Evaluate the derivative at \(x = \frac{11\pi}{6}\)

We know that \(\cos(\frac{11\pi}{6})=\frac{\sqrt{3}}{2}\) and \(\sin(\frac{11\pi}{6})=-\frac{1}{2}\).
Substitute these values into \(f^\prime(x)\):
\(f^\prime(\frac{11\pi}{6})=7\times\frac{\sqrt{3}}{2}+12\times(-\frac{1}{2})\)
\(f^\prime(\frac{11\pi}{6})=\frac{7\sqrt{3}}{2}-6\)

Answer:

\(f^\prime(x)=7\cos(x)+12\sin(x)\)
\(f^\prime(\frac{11\pi}{6})=\frac{7\sqrt{3}}{2}-6\)