To find $\frac{dy}{dx}$ for $y = \ln(x^2 + x + 4)$, we can apply the chain rule. Here's how to do it:

1. First, recognize that $y = \ln(f(x))$, where $f(x) = x^2 + x + 4$.

2. The derivative of $\ln(f(x))$ with respect to $x$ is given by:

$\frac{dy}{dx} = \frac{1}{f(x)} \cdot \frac{d}{dx}(f(x))$

3. Now, find $\frac{d}{dx}(f(x))$ for $f(x) = x^2 + x + 4$:

$\frac{d}{dx}(x^2 + x + 4) = 2x + 1$

4. Substitute $f(x)$ and its derivative into the formula:

$\frac{dy}{dx} = \frac{1}{x^2 + x + 4} \cdot (2x + 1)$

So, the derivative is:

$\frac{dy}{dx} = \frac{2x + 1}{x^2 + x + 4}$

Let me know if you'd like further explanation or details on any of the steps.

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Related Questions:

1. How would the derivative change if the function inside the logarithm was $x^3 + x + 4$?
2. What happens if you differentiate $y = \ln(2x^2 + 3x + 5)$?
3. How can the chain rule be used when differentiating logarithmic functions involving more complex terms?
4. What is the derivative of $y = e^{x^2 + x + 4}$?
5. How do logarithmic derivatives simplify certain problems involving products or quotients?

Tip: The chain rule is essential when differentiating composite functions like $y = \ln(f(x))$. Always differentiate the outer function first, and then multiply by the derivative of the inner function.