To find $h'(\pi/3)$, we need to first determine the derivative of $h(x) = \frac{\cos x}{f(x)}$ using the quotient rule.

The quotient rule states that if $h(x) = \frac{u(x)}{v(x)}$, then:

$h'(x) = \frac{v(x) u'(x) - u(x) v'(x)}{v(x)^2}$

In this case, $u(x) = \cos x$ and $v(x) = f(x)$, so we first compute $u'(x)$ and $v'(x)$:

• $u(x) = \cos x \implies u'(x) = -\sin x$
• $v(x) = f(x) \implies v'(x) = f'(x)$

Now apply the quotient rule:

$h'(x) = \frac{f(x)(-\sin x) - \cos x f'(x)}{f(x)^2}$

Substitute $x = \frac{\pi}{3}$, and use the given values $f\left(\frac{\pi}{3}\right) = -2$ and $f'\left(\frac{\pi}{3}\right) = 8$:

1. $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
2. $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$

Now substitute into the derivative formula:

$h'\left(\frac{\pi}{3}\right) = \frac{-2\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2}\right)(8)}{(-2)^2}$

Simplify:

$h'\left(\frac{\pi}{3}\right) = \frac{-\sqrt{3} - 4}{4}$

Thus, the derivative is:

$h'\left(\frac{\pi}{3}\right) = \frac{-\sqrt{3} - 4}{4}$

This is the value of $h'(\pi/3)$.

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Would you like more details or have any questions? Here are some related questions to explore:

1. How do you apply the quotient rule for more complex functions?
2. Can you verify the trigonometric values for angles like $\frac{\pi}{3}$?
3. What is the product rule, and how does it differ from the quotient rule?
4. How would the chain rule apply to functions involving trigonometric functions?
5. What happens to the behavior of $h(x)$ if $f(x)$ has zeros?

Tip: Always double-check the values of basic trigonometric functions, as they are key to simplifying problems efficiently.