Answer is: option3

\( \frac{3e}{10} \)

Solution:

\( y = x^2 \ln x \)

We need to find \( dy \) when \( x = e \) and \( dx = 0.1 \).

Differentiating both sides with respect to \( x \):

\(\frac{dy}{dx} = \frac{d}{dx} (x^2 \ln x) \).

Using the product rule, where \( u = x^2 \) and \( v = \ln x \):

\(\frac{d}{dx} (x^2 \ln x) = x^2 \cdot \frac{d}{dx} (\ln x) + \ln x \cdot \frac{d}{dx} (x^2) \).

\(= x^2 \cdot \frac{1}{x} + \ln x \cdot (2x) \).

\(= x + 2x \ln x \).

Thus,

\( dy = (x + 2x \ln x) dx \).

Substituting \( x = e \):

\( dy = (e + 2e \ln e) dx \).

Since \( \ln e = 1 \), this simplifies to:

\( dy = (e + 2e \cdot 1) dx \).

\( dy = (e + 2e) dx \).

\( dy = 3e \cdot dx \).

Substituting \( dx = 0.1 \):

\( dy = 3e \times 0.1 \).

\( dy = \frac{3e}{10} \).

The answer is \( \frac{3e}{10} \).