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To find the derivative \( f'(x) \) of the function:

\[
f(x) = (3x + 1) \ln(4x + 3)
\]

we need to apply the product rule. The product rule states that if \( f(x) = u(x) \cdot v(x) \), then:

\[
f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]

### Step 1: Define \( u(x) \) and \( v(x) \)
Let:
\[
u(x) = 3x + 1 \quad \text{and} \quad v(x) = \ln(4x + 3)
\]

### Step 2: Differentiate \( u(x) \) and \( v(x) \)
\[
u'(x) = 3
\]
\[
v'(x) = \frac{d}{dx}[\ln(4x + 3)] = \frac{4}{4x + 3}
\]

### Step 3: Apply the Product Rule
\[
f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]
\[
f'(x) = 3 \cdot \ln(4x + 3) + (3x + 1) \cdot \frac{4}{4x + 3}
\]

### Step 4: Simplify the Expression
\[
f'(x) = 3 \ln(4x + 3) + \frac{4(3x + 1)}{4x + 3}
\]

### Step 5: Evaluate at \( x = 0 \)
\[
f'(0) = 3 \ln(4(0) + 3) + \frac{4(3(0) + 1)}{4(0) + 3}
\]
\[
f'(0) = 3 \ln(3) + \frac{4 \cdot 1}{3}
\]
\[
f'(0) = 3 \ln(3) + \frac{4}{3}
\]

Thus, the derivative evaluated at \( x = 0 \) is:

\[
f'(0) = 3 \ln(3) + \frac{4}{3}
\]

Would you like further clarification or more details on this solution? 

### Related Questions:
1. How do you apply the product rule to functions involving trigonometric terms?
2. What is the derivative of \( \ln(5x + 2) \) using the chain rule?
3. Can you explain how to find the derivative of a function that involves division, using the quotient rule?
4. How do you simplify the expression \( \frac{4(3x + 1)}{4x + 3} \) further if needed?
5. What would be the result if we differentiated \( \ln(x) \) without the chain rule application?

### Tip:
When differentiating a product of two functions, make sure to clearly identify and differentiate each part separately before applying the product rule.

Find the derivative of the function f(x) = (3x + 1)ln(4x + 3) and evaluate it at x = 0.

Learn how to differentiate the function f(x) = (3x + 1)ln(4x + 3) using the product and chain rules. This step-by-step solution also includes the evaluation of the derivative at x = 0. Ideal for students studying undergraduate calculus or AP calculus.

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