Answer is: option3

Solution:

We are given the functions: \[ y = u + 2e^u \] \[ u = 1 + \ln x \] Differentiating \( y \) with respect to \( u \): \[ \frac{dy}{du} = \frac{d}{du} (u + 2e^u) \] \[ \frac{dy}{du} = 1 + 2e^u \] Differentiating \( u \) with respect to \( x \): \[ \frac{du}{dx} = \frac{1}{x} \] Using the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \] \[ \frac{dy}{dx} = (1 + 2e^u) \cdot \frac{1}{x} \] Find \( u \) at \( x = \frac{1}{e} \): \[ u = 1 + \ln \left( \frac{1}{e} \right) \] \[ u = 1 + (-1) = 0 \] Now, substitute \( u = 0 \) and \( x = \frac{1}{e} \): \[ \frac{dy}{dx} = (1 + 2e^0) \cdot \frac{1}{(1/e)} \] \[ = (1 + 2 \cdot 1) \cdot e \] \[ = (1 + 2)e \] \[ = 3e \] Final Answer: \[ \boxed{3e} \]