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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:

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

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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.

Learn how to find the derivative of the logarithmic function f(x) = ln(2x^2 - 3x + 4) using the chain rule. This solution covers key calculus concepts including derivatives and logarithmic functions, suitable for college-level calculus students.

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