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To solve this problem, we'll need to find the derivative of the function \( f(x) = (3x + 1) \ln(4x + 5) \) using the product rule, and then evaluate it at \( x = 0 \).

### Step 1: Apply the Product Rule
The product rule states:
\[
(fg)' = f'g + fg'
\]
For \( f(x) = (3x+1) \) and \( g(x) = \ln(4x+5) \):

- Let \( u = 3x+1 \) and \( v = \ln(4x+5) \).
- \( u' = 3 \).
- \( v' = \frac{4}{4x+5} \) (using the derivative of \( \ln(x) \)).

### Step 2: Differentiate \( f(x) \)
Using the product rule:
\[
f'(x) = u'v + uv'
\]
Substitute the derivatives:
\[
f'(x) = 3 \ln(4x+5) + \frac{(3x+1) \cdot 4}{4x+5}
\]

### Step 3: Simplify the Expression
\[
f'(x) = 3 \ln(4x+5) + \frac{4(3x+1)}{4x+5}
\]

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

Thus, the derivative at \( x = 0 \) is:
\[
f'(0) = 3 \ln(5) + \frac{4}{5}
\]

Would you like to go into more detail about any of these steps, or have any questions?

### Related Questions
1. How does the product rule for differentiation work in general?
2. What is the derivative of \( \ln(x) \) and why?
3. How can we apply the chain rule when differentiating logarithmic functions?
4. What is the significance of evaluating a derivative at a specific point?
5. How can derivatives be used to find the slope of a tangent line?

### Tip
When differentiating products involving logarithmic functions, it's crucial to identify the roles of each part and use the product rule carefully.

Find the derivative of the function f(x) = (3x + 1) ln(4x + 5) and then evaluate it at x = 0.

Learn how to find the derivative of f(x) = (3x + 1) ln(4x + 5) using the product rule and logarithmic differentiation. Detailed step-by-step solution for students in Grades 11-12.

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