We are tasked with finding the derivative of the expression $\frac{h(x)}{x}$ evaluated at $x = 2$, given that $h(2) = 2$ and $h'(2) = -6$.

Step-by-Step Solution:

We need to calculate:

$\left. \frac{d}{dx} \left( \frac{h(x)}{x} \right) \right|_{x=2}$

Using the quotient rule for differentiation, where:

$\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$

Here, $f(x) = h(x)$ and $g(x) = x$, so we get:

$\frac{d}{dx} \left( \frac{h(x)}{x} \right) = \frac{h'(x) \cdot x - h(x) \cdot 1}{x^2}$

Simplifying:

$\frac{d}{dx} \left( \frac{h(x)}{x} \right) = \frac{h'(x) \cdot x - h(x)}{x^2}$

Now, substitute $x = 2$, $h(2) = 2$, and $h'(2) = -6$:

$\frac{d}{dx} \left( \frac{h(x)}{x} \right)\Bigg|_{x=2} = \frac{(-6) \cdot 2 - 2}{2^2}$

Simplifying further:

$= \frac{-12 - 2}{4} = \frac{-14}{4} = -\frac{7}{2}$

Final Answer:

$\frac{d}{dx} \left( \frac{h(x)}{x} \right)\Bigg|_{x=2} = -\frac{7}{2}$

Would you like any further details or clarifications?

Here are 5 related questions to deepen your understanding:

1. How does the quotient rule apply to functions involving polynomials?
2. What is the physical interpretation of the derivative of a quotient in real-world problems?
3. How would the problem change if we were asked to differentiate $\frac{h(x)}{x^2}$ instead?
4. Can you find the second derivative of $\frac{h(x)}{x}$ at $x = 2$?
5. What other rules of differentiation (e.g., product or chain rule) could be used to solve similar problems?

Tip: Always check if you can simplify the expression before applying the quotient rule, as it can make differentiation easier.