Question

if $f(x)=sec(4x)$, then $fleft(\frac{pi}{6}
ight)=$
$-\frac{8sqrt{3}}{3}$
$\frac{2sqrt{3}}{3}$
$\frac{8sqrt{3}}{3}$
$\frac{sqrt{3}}{2}$
question 6 of 26
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Explanation:

Step1: Recall derivative of sec(u)

The derivative of $y = \sec(u)$ with respect to $x$ is $y'=\sec(u)\tan(u)\cdot u'$ by the chain - rule. Here $u = 4x$, so $u'=4$. Then $f'(x)=\sec(4x)\tan(4x)\cdot4 = 4\sec(4x)\tan(4x)$.

Step2: Substitute $x = \frac{\pi}{6}$

First, find the value of $\sec(4x)$ and $\tan(4x)$ when $x=\frac{\pi}{6}$.
When $x = \frac{\pi}{6}$, $4x=\frac{4\pi}{6}=\frac{2\pi}{3}$.
We know that $\sec(\frac{2\pi}{3})=\frac{1}{\cos(\frac{2\pi}{3})}=- 2$ and $\tan(\frac{2\pi}{3})=-\sqrt{3}$.

Step3: Calculate $f'(\frac{\pi}{6})$

Substitute $\sec(4\cdot\frac{\pi}{6})=-2$ and $\tan(4\cdot\frac{\pi}{6})=-\sqrt{3}$ into $f'(x)$.
$f'(\frac{\pi}{6})=4\sec(\frac{2\pi}{3})\tan(\frac{2\pi}{3})$.
$f'(\frac{\pi}{6})=4\times(-2)\times(-\sqrt{3}) = 8\sqrt{3}$.

Answer:

$8\sqrt{3}$