The expression in the image is:

$f(x) = \log(3x^2 + 9x - 13)$

To differentiate this function with respect to $x$, we'll apply the chain rule.

Step-by-Step Differentiation:

1. Outer Function: The outer function is the logarithm, $\log(u)$, where $u = 3x^2 + 9x - 13$.

   The derivative of $\log(u)$ with respect to $u$ is: $\frac{d}{du} \log(u) = \frac{1}{u}$

2. Inner Function: The inner function is $u = 3x^2 + 9x - 13$.

   The derivative of $u$ with respect to $x$ is: $\frac{du}{dx} = \frac{d}{dx} (3x^2 + 9x - 13) = 6x + 9$

3. Applying the Chain Rule: $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = \frac{1}{3x^2 + 9x - 13} \cdot (6x + 9)$

Final Answer:

$f'(x) = \frac{6x + 9}{3x^2 + 9x - 13}$

Would you like further details or have any questions about this differentiation process?

Related Questions:

1. What are the critical points of $f(x) = \log(3x^2 + 9x - 13)$?
2. How would you find the second derivative of $f(x)$?
3. What is the domain of $f(x)$?
4. How can you determine the concavity of $f(x)$ using the second derivative?
5. Can you integrate $f(x)$? If so, what is the integral?

Tip:

When differentiating functions involving logarithms, always remember to apply the chain rule, starting from the outermost function.