Differentiate. \[ g(x)=(x+9 \sqrt{x}) e^{x} \] \( g^{\prime}(x)=\square \)

To differentiate the function \( g(x) = (x + 9\sqrt{x}) e^{x} \), we can use the product rule, which states that \( (uv)' = u'v + uv' \) where \( u = (x + 9\sqrt{x}) \) and \( v = e^{x} \). First, we differentiate \( u \): \[ u' = 1 + \frac{9}{2\sqrt{x}}. \] Now, differentiate \( v \): \[ v' = e^{x}. \] Applying the product rule gives us: \[ g'(x) = (1 + \frac{9}{2\sqrt{x}}) e^{x} + (x + 9\sqrt{x}) e^{x}. \] We can factor \( e^{x} \) out: \[ g'(x) = e^{x} \left(1 + \frac{9}{2\sqrt{x}} + x + 9\sqrt{x}\right). \] Thus, simplifying inside the parentheses, we arrive at: \[ g'(x) = e^{x} \left(x + 9\sqrt{x} + 1 + \frac{9}{2\sqrt{x}}\right). \] Finally, we have: \[ g^{\prime}(x) = e^{x} \left(x + 9\sqrt{x} + 1 + \frac{9}{2\sqrt{x}}\right). \]