We are asked to find $g' \left( \frac{\pi}{3} \right)$ given that:

$g(x) = f(x) \sin(x)$ with additional information: $f \left( \frac{\pi}{3} \right) = 4 \quad \text{and} \quad f' \left( \frac{\pi}{3} \right) = -3$

Step 1: Use the product rule to differentiate $g(x)$

The function $g(x) = f(x) \sin(x)$ is a product of two functions, $f(x)$ and $\sin(x)$. Using the product rule:

$g'(x) = f'(x) \sin(x) + f(x) \cos(x)$

Step 2: Plug in $x = \frac{\pi}{3}$

Now we substitute $x = \frac{\pi}{3}$ into the derivative:

$g' \left( \frac{\pi}{3} \right) = f' \left( \frac{\pi}{3} \right) \sin \left( \frac{\pi}{3} \right) + f \left( \frac{\pi}{3} \right) \cos \left( \frac{\pi}{3} \right)$

We know that: $f \left( \frac{\pi}{3} \right) = 4 \quad \text{and} \quad f' \left( \frac{\pi}{3} \right) = -3$

Also, from trigonometry: $\sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \cos \left( \frac{\pi}{3} \right) = \frac{1}{2}$

Step 3: Calculate the values

Substitute these into the expression for $g' \left( \frac{\pi}{3} \right)$:

$g' \left( \frac{\pi}{3} \right) = (-3) \cdot \frac{\sqrt{3}}{2} + 4 \cdot \frac{1}{2}$

This simplifies to:

$g' \left( \frac{\pi}{3} \right) = -\frac{3\sqrt{3}}{2} + 2$

Thus, the final answer is:

$g' \left( \frac{\pi}{3} \right) = 2 - \frac{3\sqrt{3}}{2}$

Would you like further details or have any questions on this?

Here are some related questions:

1. How would the derivative change if $g(x) = f(x) \cos(x)$?
2. What is the second derivative of $g(x) = f(x) \sin(x)$?
3. How do we apply the quotient rule to the function $h(x)$?
4. What would $g' \left( \frac{\pi}{2} \right)$ be if $f(x)$ remains the same?
5. How does the value of $f(x)$ affect the shape of the function $g(x)$?

Tip: Always apply the product rule carefully, ensuring that each term is properly differentiated and evaluated at the given point.