If f(x) = x^3log_e x , then f''(e^2) is equal to

The correct answer is Option (2) → \(17e^2\)

By the product rule:

    \( \displaystyle f'(x) = 3x^2\log_e x + x^3 \cdot \frac{1}{x} = 3x^2\log_e x + x^2 \)

Differentiating again, with the product rule on the first term:

    \( \displaystyle f''(x) = 3\left(2x\log_e x + x^2 \cdot \frac{1}{x}\right) + 2x \)

    \(f''(x) = 6x\log_e x + 3x + 2x = 6x\log_e x + 5x\)

At \(x = e^2\), \(\log_e e^2 = 2\):

    \(f''(e^2) = 6e^2(2) + 5e^2 = 12e^2 + 5e^2 = 17e^2\)