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 the derivative at $x = \frac{\pi}{6}$:

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

Calculating inside the cosine:

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

Thus,

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

Using the property that $\cos(-x) = \cos(x)$:

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

Now, we know that $\cos\left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$:

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