If \(y = x + \displaystyle\frac{1}{x}\), \(x \gt 0\), then the value of \(\displaystyle\frac{d^{2}y}{dx^{2}}\) at \(x = 2\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{4}\)
Writing \(\frac{1}{x} = x^{-1}\) and using the power rule:
\( \displaystyle \frac{dy}{dx} = 1 - x^{-2} = 1 - \frac{1}{x^{2}} \)
Differentiating again for the second derivative:
\( \displaystyle \frac{d^{2}y}{dx^{2}} = 2x^{-3} = \frac{2}{x^{3}} \)
At x = 2:
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \frac{2}{8} = \frac{1}{4} \)