To solve for $f'(x)$ when $f(x) = \ln(2x^2 - 3x + 4)$, we will apply the chain rule.

Solution:

Given: $f(x) = \ln(2x^2 - 3x + 4)$

Using the derivative formula for $\ln(u)$: $\frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot u'$

Here, $u = 2x^2 - 3x + 4$. Now, we differentiate $u$ with respect to $x$:

$u' = \frac{d}{dx}(2x^2 - 3x + 4) = 4x - 3$

Now, substitute into the derivative formula:

$f'(x) = \frac{1}{2x^2 - 3x + 4} \cdot (4x - 3)$

Thus, the derivative is:

$f'(x) = \frac{4x - 3}{2x^2 - 3x + 4}$

Do you want more details or have any questions?

Here are five related questions you might explore:

1. What is the domain of the function $f(x) = \ln(2x^2 - 3x + 4)$?
2. How do logarithmic functions behave as $x \to \infty$?
3. Can you solve for the critical points of $f(x) = \ln(2x^2 - 3x + 4)$?
4. What would the second derivative $f''(x)$ be for this function?
5. How does the graph of $f(x)$ compare to the function $g(x) = \ln(x)$?

Tip: When working with logarithmic functions, always verify that the argument of the logarithm is positive to ensure the function is well-defined.