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.