Given that $f(x) = 4 \sin \left( 3x - \frac{\pi}{3} \right)$, evaluate $f' \left( \frac{\pi}{6} \right)$. - Scottish Highers Maths - Question 12 - 2022

Next, we need to evaluate $f' \left( \frac{\pi}{6} \right)$:

First, substitute $x = \frac{\pi}{6}$ into the derivative:

$f' \left( \frac{\pi}{6} \right) = 12 \cos \left( 3 \cdot \frac{\pi}{6} - \frac{\pi}{3} \right)$

This simplifies to:

$= 12 \cos \left( \frac{\pi}{2} - \frac{\pi}{3} \right)$

Calculating the angle gives us:

$= 12 \cos \left( \frac{\pi}{2} - \frac{\pi}{3} \right) = 12 \cos \left( \frac{\pi}{6} \right)$

We know that:

$\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$

Thus we have:

$f' \left( \frac{\pi}{6} \right) = 12 \cdot \frac{\sqrt{3}}{2} = 6\sqrt{3}.$