To find the derivative of \( \frac{5\ln(x)}{4x + 3} \) and then evaluate it at \( x = e \), let's go through the steps:

### Step 1: Apply the quotient rule for derivatives.
The function \( f(x) = \frac{u(x)}{v(x)} \) where:
- \( u(x) = 5\ln(x) \)
- \( v(x) = 4x + 3 \)

The quotient rule states:
\[
f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}
\]

### Step 2: Compute \( u'(x) \) and \( v'(x) \).
- \( u(x) = 5\ln(x) \) so \( u'(x) = \frac{5}{x} \)
- \( v(x) = 4x + 3 \) so \( v'(x) = 4 \)

### Step 3: Apply the quotient rule.
Substituting \( u'(x) \), \( u(x) \), \( v(x) \), and \( v'(x) \) into the quotient rule formula:
\[
f'(x) = \frac{\left(\frac{5}{x}\right)(4x + 3) - 5\ln(x)(4)}{(4x + 3)^2}
\]

### Step 4: Simplify the derivative.
Simplify the numerator:
\[
f'(x) = \frac{\frac{5(4x + 3)}{x} - 20\ln(x)}{(4x + 3)^2}
\]
Now, evaluate this expression at \( x = e \).

### Step 5: Evaluate at \( x = e \).
For \( x = e \):
- \( \ln(e) = 1 \)
- The expression simplifies to:
  \[
  f'(e) = \frac{\frac{5(4e + 3)}{e} - 20(1)}{(4e + 3)^2}
  \]
Simplifying further:
\[
f'(e) = \frac{\frac{5(4e + 3)}{e} - 20}{(4e + 3)^2} = \frac{\frac{5(4e + 3)}{e} - 20}{(4e + 3)^2}
\]
We can now compute this expression numerically.

Let me calculate the result.The derivative of \( \frac{5\ln(x)}{4x + 3} \) evaluated at \( x = e \) is approximately \( 0.0287 \).

Would you like further details or clarification on any step?

Here are 5 related questions to expand on this concept:
1. How does the quotient rule apply to other types of functions?
2. What is the general process of evaluating a derivative at a specific point?
3. How does logarithmic differentiation help in more complex expressions?
4. What happens when we evaluate the second derivative of this function?
5. Can we simplify this function before differentiating to make the process easier?

**Tip**: When working with natural logarithms, remember that \( \ln(e) = 1 \), which can often simplify your calculations!

This page explains how to find the derivative of the function 5ln(x) / (4x + 3) using the quotient rule and evaluates the derivative at x = e. This problem covers key calculus concepts, including logarithmic differentiation and the quotient rule, suitable for high school students in Grades 11-12.

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