If \(y=x^2\log_e x\), then \(\displaystyle\frac{d^2y}{dx^2}\) is equal to :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(3+2\log_e x\)
By the product rule:
\( \displaystyle \frac{dy}{dx} = 2x\log_e x + x^2\cdot\frac{1}{x} = 2x\log_e x + x \)
Differentiating again:
\( \displaystyle \frac{d^2y}{dx^2} = 2\log_e x + 2x\cdot\frac{1}{x} + 1 \)
\( = 2\log_e x + 3 \)
Why the other options are incorrect
Option (1) \(5+x\log_e x\) — Incorrect. The second derivative is \(3 + 2\log_e x\); it has no \(x\log_e x\) term.
Option (3) \(3+2x\log_e x\) — Incorrect. The derivative of \(2x\log_e x\) is \(2\log_e x + 2\); the log term loses its x.
Option (4) \(x(1+2\log_e x)\) — Incorrect. This equals \(x + 2x\log_e x\), the first derivative, not the second.