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

Now, we need to substitute ( x = \frac{\pi}{6} ) into the derivative:

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

Calculating the argument of the cosine:

$3 \cdot \frac{\pi}{6} = \frac{\pi}{2} \quad \Rightarrow \quad \frac{\pi}{2} - \frac{\pi}{3} = \frac{3\pi - 2\pi}{6} = \frac{\pi}{6}$

Now we have:

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

Knowing that ( \cos \left( \frac{\pi}{6} ight) = \frac{\sqrt{3}}{2} ), we can substitute:

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

Thus, the final answer is:

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