To find $f'(-3)$ for the function $f(x) = \frac{1}{x - 3}$, we will use the definition of the derivative:

$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

In this case, $a = -3$, so the expression for $f'(-3)$ becomes:

$f'(-3) = \lim_{x \to -3} \frac{f(x) - f(-3)}{x + 3}$

Step 1: Compute $f(-3)$

First, we need to evaluate $f(-3)$:

$f(-3) = \frac{1}{-3 - 3} = \frac{1}{-6} = -\frac{1}{6}$

Step 2: Set up the limit

Now substitute $f(x) = \frac{1}{x - 3}$ and $f(-3) = -\frac{1}{6}$ into the limit definition of the derivative:

$f'(-3) = \lim_{x \to -3} \frac{\frac{1}{x - 3} - \left( -\frac{1}{6} \right)}{x + 3}$

This simplifies to:

$f'(-3) = \lim_{x \to -3} \frac{\frac{1}{x - 3} + \frac{1}{6}}{x + 3}$

Step 3: Combine the fractions

To combine the fractions in the numerator, we need a common denominator. The common denominator is $6(x - 3)$, so we rewrite the numerator as:

$\frac{1}{x - 3} + \frac{1}{6} = \frac{6 + (x - 3)}{6(x - 3)} = \frac{x + 3}{6(x - 3)}$

Step 4: Simplify the expression

Now substitute this back into the limit:

$f'(-3) = \lim_{x \to -3} \frac{\frac{x + 3}{6(x - 3)}}{x + 3}$

This simplifies to:

$f'(-3) = \lim_{x \to -3} \frac{1}{6(x - 3)}$

Step 5: Evaluate the limit

As $x \to -3$, $x - 3 \to -6$. So we have:

$f'(-3) = \frac{1}{6(-6)} = \frac{1}{-36} = -\frac{1}{36}$

Final Answer:

$f'(-3) = -\frac{1}{36}$

Would you like further details on any step, or have any questions?

Here are 5 related questions to deepen understanding:

1. How is the derivative related to the slope of the tangent line?
2. Why do we use limits to find the derivative?
3. How does the derivative of $f(x) = \frac{1}{x-3}$ change as you vary the point $x = a$?
4. What does the negative slope tell you about the function at $x = -3$?
5. Can you apply the same method to find the derivative of $f(x) = \frac{1}{x+2}$?

Tip: When combining fractions, always look for a common denominator to simplify the expression effectively!