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To solve this problem using the quotient rule, we need to apply the formula for the derivative of a quotient of two functions:

\[
\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}
\]

Given that:

\[
f(x) = \frac{6x - 5}{2x + 1}
\]

Here, 
- \( u(x) = 6x - 5 \)
- \( v(x) = 2x + 1 \)

Now, let's find the derivatives of both \( u(x) \) and \( v(x) \):
- \( u'(x) = 6 \)
- \( v'(x) = 2 \)

Now, applying the quotient rule:

\[
f'(x) = \frac{(6)(2x + 1) - (6x - 5)(2)}{(2x + 1)^2}
\]

Let's simplify this expression step by step:

1. First, compute the products:
   - \( (6)(2x + 1) = 12x + 6 \)
   - \( (6x - 5)(2) = 12x - 10 \)

2. Now, subtract the second term from the first:
   \[
   12x + 6 - (12x - 10) = 12x + 6 - 12x + 10 = 16
   \]

3. Therefore, the derivative is:
   \[
   f'(x) = \frac{16}{(2x + 1)^2}
   \]

So, the derivative of the function is:

\[
f'(x) = \frac{16}{(2x + 1)^2}
\]

Would you like more details on any step of this process?

Here are 5 related questions to expand on this concept:

1. How does the quotient rule compare to the product rule in calculus?
2. Can you explain the conditions where the quotient rule might fail or be unnecessary?
3. How would the result differ if the denominator had a higher degree than the numerator?
4. What happens to the derivative when both the numerator and the denominator are linear functions?
5. How can we check the correctness of the derivative using an alternative method?

**Tip:** Always ensure that the denominator in the quotient rule is non-zero to avoid division by zero errors.

Use the quotient rule to find the derivative of the function: f(x) = (6x - 5) / (2x + 1)

Learn how to use the quotient rule to find the derivative of the function f(x) = (6x - 5) / (2x + 1). This calculus problem covers key concepts like the quotient rule for derivatives. Suitable for high school and early college-level calculus students.

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