​$$\text{If } y = u^2 + \log u \text{ and } u = e^x, \text{ then find } \frac{dy}{dx}:$$​​

Correct option is B

​$$\text{To differentiate } y \text{ with respect to } x, \text{ we use the chain rule:} \\\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

$$\text{First, differentiate } y \text{ with respect to } u: \\\frac{dy}{du} = 2u + \frac{1}{u} \\\text{Then, differentiate } u = e^x \text{ with respect to } x: \\\frac{du}{dx} = e^x \\\text{Now, apply the chain rule:} \\\frac{dy}{dx} = \left( 2u + \frac{1}{u} \right) \cdot \frac{du}{dx} = \left( 2e^x + \frac{1}{e^x} \right) \cdot e^x \\\text{Simplifying:} \\\frac{dy}{dx} = 2e^{2x} + 1$$​​​