To differentiate the function $y = 7e^x + 2\ln(x)$, we will apply the rules for differentiation.

1. Differentiate $7e^x$:

   • The derivative of $e^x$ is $e^x$.
   • So, $\frac{d}{dx}(7e^x) = 7e^x$.

2. Differentiate $2\ln(x)$:

   • The derivative of $\ln(x)$ is $\frac{1}{x}$.
   • So, $\frac{d}{dx}(2\ln(x)) = 2 \cdot \frac{1}{x} = \frac{2}{x}$.

Now, combining both results:

$y' = 7e^x + \frac{2}{x}$

Final Answer:

$y' = 7e^x + \frac{2}{x}$

Would you like further details or have any questions?

Here are some related questions to explore further:

1. How would you differentiate $y = e^{3x} + \ln(x^2)$?
2. What is the derivative of $y = 5e^{x^2}$?
3. How would you differentiate a function involving $\ln(x)$ in the denominator?
4. How do the product and chain rules apply when differentiating exponential functions?
5. How do you find the second derivative of $y = 7e^x + 2\ln(x)$?

Tip: For logarithmic functions, remember that the derivative of $\ln(x)$ is $\frac{1}{x}$, which is especially useful when combined with constants or more complex expressions.