To find the derivative of the function

$f(x) = \frac{1}{3 - 2x}$

at the point $(-1, \frac{1}{5})$, we can follow these steps:

Step 1: Differentiate the function

Using the chain rule, the derivative of $f(x)$ can be found. First, rewrite the function as: $f(x) = (3 - 2x)^{-1}$

Now, apply the chain rule:

$f'(x) = -1 \cdot (3 - 2x)^{-2} \cdot (-2) = \frac{2}{(3 - 2x)^2}$

Step 2: Evaluate the derivative at $x = -1$

Now substitute $x = -1$ into the derivative formula:

$f'(-1) = \frac{2}{(3 - 2(-1))^2} = \frac{2}{(3 + 2)^2} = \frac{2}{5^2} = \frac{2}{25}$

Conclusion:

The derivative of the function at the point $(-1, \frac{1}{5})$ is:

$f'(-1) = \frac{2}{25}$

Let me know if you'd like further details or clarifications.

Here are 5 related questions:

1. How would the derivative change if the denominator was different, e.g., $3 + 2x$?
2. What is the second derivative of $f(x)$?
3. How do you interpret the value of the derivative geometrically at the point $(-1, \frac{1}{5})$?
4. What happens to the derivative as $x$ approaches infinity?
5. How does the chain rule apply to more complex functions?

Tip: When using the chain rule, remember that you differentiate the outer function and multiply by the derivative of the inner function.