Let's begin solving part (a) of the problem using the formal definition of the derivative.

The given function is: $f(x) = \frac{6}{x}$

The formal definition of the derivative is: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

We are tasked with finding the derivative at $x = 5$, so we need to compute $f'(5)$: $f'(5) = \lim_{h \to 0} \frac{f(5+h) - f(5)}{h}$

First, let's substitute $f(x) = \frac{6}{x}$ into this limit:

$f'(5) = \lim_{h \to 0} \frac{\frac{6}{5+h} - \frac{6}{5}}{h}$

Now let's simplify the expression inside the limit: $f'(5) = \lim_{h \to 0} \frac{6 \left(\frac{1}{5+h} - \frac{1}{5}\right)}{h}$

To combine the terms inside the parentheses, we find a common denominator: $\frac{1}{5+h} - \frac{1}{5} = \frac{5 - (5+h)}{5(5+h)} = \frac{-h}{5(5+h)}$

Substituting this into the limit expression: $f'(5) = \lim_{h \to 0} \frac{6 \cdot \frac{-h}{5(5+h)}}{h}$

Simplifying: $f'(5) = \lim_{h \to 0} \frac{-6h}{h \cdot 5(5+h)}$

Cancel out the $h$ in the numerator and denominator: $f'(5) = \lim_{h \to 0} \frac{-6}{5(5+h)}$

Now take the limit as $h \to 0$: $f'(5) = \frac{-6}{5(5)} = \frac{-6}{25}$

Thus, the derivative at $x = 5$ is: $f'(5) = \frac{-6}{25}$

Would you like to continue to part (b) or ask questions? Let me know if you need more details on this solution!

Here are 5 related questions:

1. How would the derivative change if the function were $f(x) = \frac{c}{x}$ where $c$ is a constant?
2. Can you use the product rule to differentiate $f(x) = \frac{6}{x}$ instead of using the definition of the derivative?
3. What is the significance of the sign of $f'(5)$? What does it tell you about the slope of the tangent line at $x = 5$?
4. How would you find the second derivative of $f(x) = \frac{6}{x}$?
5. How would the graph of the function $y = \frac{6}{x}$ change if the numerator were a negative constant?

Tip: The derivative at a point gives you the slope of the tangent line to the curve at that point.