If \(y = \left(\displaystyle\frac{1}{x}\right)^x\), then value of \(e^e\left(\displaystyle\frac{d^2y}{dx^2}\right)_{x=e}\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(4 - \displaystyle\frac{1}{e}\)
The variable is in both the base and the power, so use logarithmic differentiation:
\( \displaystyle \log y = x\log\frac{1}{x} = -x\log x \)
Differentiating with respect to x:
\( \displaystyle \frac{1}{y}\frac{dy}{dx} = -(\log x + 1) \)
\( \displaystyle \frac{dy}{dx} = -y(1 + \log x) \qquad (1) \)
Differentiating again by the product rule:
\( \displaystyle \frac{d^2y}{dx^2} = -\frac{dy}{dx}(1 + \log x) - \frac{y}{x} \)
Putting in dy/dx from (1):
\( \displaystyle \frac{d^2y}{dx^2} = y(1 + \log x)^2 - \frac{y}{x} \)
At x = e, log e = 1 and \(y = \left(\frac{1}{e}\right)^e = e^{-e}\):
\( \displaystyle \left(\frac{d^2y}{dx^2}\right)_{x=e} = e^{-e}\left(4 - \frac{1}{e}\right) \)
Multiplying by \(e^e\):
\( \displaystyle e^e\left(\frac{d^2y}{dx^2}\right)_{x=e} = 4 - \frac{1}{e} \)