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We are asked to differentiate the function \( g(x) = (x + 2\sqrt{x}) e^x \).

Let’s apply the **product rule**, since we have two functions being multiplied:
\[
\frac{d}{dx} \left[ u(x) \cdot v(x) \right] = u'(x) \cdot v(x) + u(x) \cdot v'(x).
\]
Here:
- \( u(x) = x + 2\sqrt{x} \)
- \( v(x) = e^x \)

---

### Step 1: Differentiate \( u(x) = x + 2\sqrt{x} \)
Using the power rule:
- \( \frac{d}{dx}(x) = 1 \)
- \( \frac{d}{dx}(2\sqrt{x}) = 2 \cdot \frac{d}{dx}(x^{1/2}) = 2 \cdot \frac{1}{2} x^{-1/2} = \frac{1}{\sqrt{x}} \).

Thus:
\[
u'(x) = 1 + \frac{1}{\sqrt{x}}.
\]

---

### Step 2: Differentiate \( v(x) = e^x \)
\[
v'(x) = e^x.
\]

---

### Step 3: Apply the Product Rule
Now we substitute everything into the product rule:
\[
g'(x) = \left( 1 + \frac{1}{\sqrt{x}} \right) e^x + (x + 2\sqrt{x}) e^x.
\]

---

### Step 4: Factor out \( e^x \)
We can factor out \( e^x \) from both terms:
\[
g'(x) = e^x \left( \left( 1 + \frac{1}{\sqrt{x}} \right) + (x + 2\sqrt{x}) \right).
\]

Now simplify inside the parentheses:
\[
1 + \frac{1}{\sqrt{x}} + x + 2\sqrt{x}.
\]

Thus, the derivative is:
\[
\boxed{g'(x) = e^x \left( x + 2\sqrt{x} + 1 + \frac{1}{\sqrt{x}} \right).}
\]

---

Would you like more details, such as further simplification or examples of applying product rule? 

Here are 5 follow-up questions that may deepen your understanding:
1. How does the product rule compare to the chain rule?
2. What are the domain restrictions for \( g(x) = (x + 2\sqrt{x}) e^x \)?
3. What happens if we use the quotient rule incorrectly here?
4. How can we use this derivative to find the critical points of \( g(x) \)?
5. Can you solve for \( g'(x) = 0 \) to determine local extrema?

**Tip**: Always look for opportunities to factor the exponential \( e^x \) when differentiating functions with it—it simplifies results and makes equations easier to work with.

Learn how to differentiate the function g(x) = (x + 2√x) * e^x using the product rule. The solution includes applying the product rule, power rule, and simplifying the final result. Suitable for college calculus students.

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