If \(y = x^{2}\log_e x\), \(x \gt 0\), then \(\displaystyle\frac{d^{2}y}{dx^{2}}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2\log_e x + 3\)
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 for the second derivative:
\( \displaystyle \frac{d^{2}y}{dx^{2}} = 2\log_e x + 2x \cdot \frac{1}{x} + 1 \)
\( = 2\log_e x + 3 \)