Let's analyze the problem and solve it step by step.

Given:

• $f\left(\frac{\pi}{3}\right) = 4$
• $f'\left(\frac{\pi}{3}\right) = -3$
• $g(x) = f(x) \sin(x)$
• $h(x) = \frac{\cos(x)}{f(x)}$

You are asked to find:

• $g'\left(\frac{\pi}{3}\right)$
• $h'\left(\frac{\pi}{3}\right)$

(a) Finding $g'\left(\frac{\pi}{3}\right)$:

The function $g(x) = f(x) \sin(x)$ is a product of two functions, so we apply the product rule for differentiation: $g'(x) = f'(x) \sin(x) + f(x) \cos(x)$

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

Using the values:

• $f\left(\frac{\pi}{3}\right) = 4$
• $f'\left(\frac{\pi}{3}\right) = -3$
• $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
• $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$

Substitute these into the equation: $g'\left(\frac{\pi}{3}\right) = (-3) \cdot \frac{\sqrt{3}}{2} + 4 \cdot \frac{1}{2}$ $g'\left(\frac{\pi}{3}\right) = -\frac{3\sqrt{3}}{2} + 2$

Thus, the answer is: $g'\left(\frac{\pi}{3}\right) = 2 - \frac{3\sqrt{3}}{2}$

(b) Finding $h'\left(\frac{\pi}{3}\right)$:

The function $h(x) = \frac{\cos(x)}{f(x)}$ is a quotient, so we apply the quotient rule for differentiation: $h'(x) = \frac{f(x) \cdot (-\sin(x)) - \cos(x) \cdot f'(x)}{[f(x)]^2}$

Now substitute $x = \frac{\pi}{3}$: $h'\left(\frac{\pi}{3}\right) = \frac{f\left(\frac{\pi}{3}\right) \cdot (-\sin\left(\frac{\pi}{3}\right)) - \cos\left(\frac{\pi}{3}\right) \cdot f'\left(\frac{\pi}{3}\right)}{[f\left(\frac{\pi}{3}\right)]^2}$

Substituting the known values: $h'\left(\frac{\pi}{3}\right) = \frac{4 \cdot \left(-\frac{\sqrt{3}}{2}\right) - \frac{1}{2} \cdot (-3)}{4^2}$ $h'\left(\frac{\pi}{3}\right) = \frac{-2\sqrt{3} + \frac{3}{2}}{16}$ $h'\left(\frac{\pi}{3}\right) = \frac{-2\sqrt{3} + 1.5}{16}$

Thus, the answer is: $h'\left(\frac{\pi}{3}\right) = \frac{-2\sqrt{3} + 1.5}{16}$

Final Answers:

• $g'\left(\frac{\pi}{3}\right) = 2 - \frac{3\sqrt{3}}{2}$
• $h'\left(\frac{\pi}{3}\right) = \frac{-2\sqrt{3} + 1.5}{16}$

Do you want further details on any step?
Here are some related questions for deeper understanding:

1. How does the product rule for differentiation apply in this case?
2. Can you derive the quotient rule from first principles?
3. What is the significance of the trigonometric values at $\frac{\pi}{3}$?
4. Could we have used any alternative method to differentiate these functions?
5. How would the results change if different values were given for $f(x)$ and $f'(x)$?

Tip: For complex derivatives, always write down the rules (product, quotient, etc.) before substituting values to avoid mistakes.