To find the derivative of the function

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

at the point $(-1, \frac{1}{5})$ without using the chain rule, we can rewrite the function and apply the quotient rule. Here's how:

Step 1: Rewrite the function

We have the function:

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

This can be seen as a quotient:

$f(x) = \frac{u(x)}{v(x)}$

where $u(x) = 1$ (constant) and $v(x) = 3 - 2x$.

Step 2: Apply the quotient rule

The quotient rule states that for a function $f(x) = \frac{u(x)}{v(x)}$, the derivative is:

$f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}$

• $u(x) = 1$, so $u'(x) = 0$ (since the derivative of a constant is 0).
• $v(x) = 3 - 2x$, so $v'(x) = -2$ (the derivative of $3 - 2x$).

Now substitute these values into the quotient rule formula:

$f'(x) = \frac{(3 - 2x)(0) - (1)(-2)}{(3 - 2x)^2}$

Simplify:

$f'(x) = \frac{0 + 2}{(3 - 2x)^2} = \frac{2}{(3 - 2x)^2}$

Step 3: Evaluate the derivative at $x = -1$

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

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

Conclusion:

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

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

Would you like further explanations?

Here are 5 related questions:

1. What is the quotient rule and how is it derived?
2. How do you find the derivative of more complex rational functions?
3. How can you verify the result using a different method like implicit differentiation?
4. Can you compute the derivative of a function without any rules, using the limit definition?
5. How does the behavior of the function $f(x)$ change as $x$ approaches values where the denominator tends to zero?

Tip: The quotient rule is especially useful when the function is expressed as a ratio of two other functions, and it avoids needing to rewrite the function using negative exponents.