The given problem asks:

"If $f(x) = \ln(3x^2 + x)$, find $f'(1)$."

Let’s solve it step-by-step.

Step 1: Differentiate $f(x) = \ln(3x^2 + x)$ using the chain rule.

The chain rule states: $\frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx}$ Here, $u = 3x^2 + x$. So, first, differentiate $u$ with respect to $x$.

$\frac{du}{dx} = \frac{d}{dx}(3x^2 + x) = 6x + 1$

Now apply the chain rule to find $f'(x)$:

$f'(x) = \frac{1}{3x^2 + x} \cdot (6x + 1)$

Step 2: Evaluate $f'(x)$ at $x = 1$.

Substitute $x = 1$ into the derivative expression:

$f'(1) = \frac{6(1) + 1}{3(1)^2 + 1} = \frac{6 + 1}{3 + 1} = \frac{7}{4}$

Final Answer:

$f'(1) = \frac{7}{4}$

Would you like further details, or do you have other questions?

Here are five related questions:

1. How does the chain rule work in more complex functions like $\ln(\sin(x^2))$?
2. What is the significance of evaluating a derivative at a specific point?
3. How would the derivative change if the argument inside the logarithm was $3x^2 + x + 1$?
4. What are the domain restrictions for the function $f(x) = \ln(3x^2 + x)$?
5. Can you apply implicit differentiation to logarithmic functions?

Tip: Always ensure the argument of a logarithmic function is positive to avoid undefined values.