Let's analyze the problem step by step.

Given the function $f(x) = \frac{7}{x - 6}$, we are asked to:

Part (a): Use the limit definition of the derivative to find $f'(x)$.

The limit definition of the derivative is:

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

For the function $f(x) = \frac{7}{x - 6}$, let's plug this into the formula:

$f'(x) = \lim_{h \to 0} \frac{\frac{7}{(x + h) - 6} - \frac{7}{x - 6}}{h}$

Simplify the numerator:

$\frac{7}{(x + h) - 6} - \frac{7}{x - 6} = 7 \left( \frac{1}{(x + h) - 6} - \frac{1}{x - 6} \right)$

This can be further simplified by finding a common denominator:

$= 7 \left( \frac{(x - 6) - ((x + h) - 6)}{((x + h) - 6)(x - 6)} \right)$

Simplify the numerator of this expression:

$= 7 \left( \frac{x - 6 - x - h + 6}{((x + h) - 6)(x - 6)} \right) = 7 \left( \frac{-h}{(x - 6)(x + h - 6)} \right)$

Now, substitute this back into the limit definition of the derivative:

$f'(x) = \lim_{h \to 0} \frac{7 \left( \frac{-h}{(x - 6)(x + h - 6)} \right)}{h}$

Simplify this expression:

$f'(x) = \lim_{h \to 0} \frac{-7h}{h (x - 6)(x + h - 6)} = \lim_{h \to 0} \frac{-7}{(x - 6)(x + h - 6)}$

As $h \to 0$, the term $x + h - 6$ approaches $x - 6$, so the derivative becomes:

$f'(x) = \frac{-7}{(x - 6)^2}$

Part (b): Find the equation of the tangent line at $x = 7$.

To find the equation of the tangent line, we need the slope of the tangent line at $x = 7$ and the point $(7, f(7))$.

The slope of the tangent line is given by $f'(7)$:

$f'(7) = \frac{-7}{(7 - 6)^2} = \frac{-7}{1^2} = -7$

The point on the function at $x = 7$ is:

$f(7) = \frac{7}{7 - 6} = \frac{7}{1} = 7$

Now, we have the point $(7, 7)$ and the slope $-7$. Using the point-slope form of the equation of a line:

$y - y_1 = m(x - x_1)$

Substitute $(x_1, y_1) = (7, 7)$ and $m = -7$:

$y - 7 = -7(x - 7)$

Simplify this equation:

$y - 7 = -7x + 49$

$y = -7x + 56$

Thus, the equation of the tangent line at $x = 7$ is:

$y = -7x + 56$

Would you like more details or clarifications on any of these steps?

Here are five related questions to deepen your understanding:

1. How do we interpret the result of $f'(x) = \frac{-7}{(x - 6)^2}$ in terms of the behavior of the function $f(x)$?
2. What happens to the slope of the tangent line as $x$ approaches $6$?
3. Why do we use the point-slope form to find the equation of the tangent line?
4. What would the tangent line look like at other points on the curve, such as $x = 8$?
5. How would we use the chain rule to differentiate more complicated rational functions?

Tip: The limit definition of the derivative is essential for understanding how calculus generalizes the concept of slope from lines to curves. Practicing it helps solidify your intuition for changes in rates.