To solve for $f'(x)$, where $f(x) = 2 \sin(x) + \ln(5x)$, we need to take the derivative of each term individually.

1. Derivative of $2 \sin(x)$: $\frac{d}{dx}[2 \sin(x)] = 2 \cos(x)$

2. Derivative of $\ln(5x)$: Using the chain rule, the derivative of $\ln(5x)$ is: $\frac{d}{dx}[\ln(5x)] = \frac{1}{5x} \cdot 5 = \frac{1}{x}$

So, combining these, the derivative $f'(x)$ is: $f'(x) = 2 \cos(x) + \frac{1}{x}$

Would you like further details on the graph or have any questions?

Here are some related questions to explore further:

1. What is the second derivative $f''(x)$ of the function?
2. How would the graph of $f(x)$ and $f'(x)$ differ visually?
3. How does the term $\ln(5x)$ affect the behavior of the derivative?
4. What are the critical points of $f'(x)$?
5. How do we find the inflection points of $f(x)$?

Tip: When applying the chain rule, always check if constants inside the logarithm or trigonometric functions need to be factored out!